Additive combinatorics studies how sets behave when their elements are added, and what the resulting sums reveal about the original sets. Its gold standard is a precise chain from assumptions to evidence: specify the number system, distinguish distinct sums from repeated representations, establish a valid bound, and explain exactly what structure that bound supports.
Two collections can contain the same number of values yet produce very different numbers of totals. A regularly spaced collection reuses many totals; another collection may produce almost a different total for every unordered pair. This guide makes that contrast concrete, then develops the tools needed to recognise useful structure without mistaking one small example for a general theorem.
The Gold Standard master connects this guide to the wider subject library. The Education parent explains the standard of learning that survives a changed question. For other mechanisms, start at the How X Works hub.
Choose your additive combinatorics route
Chapters 1–3: Sets, sums and basic bounds
Chapters 4–6: Progressions, collisions and probability
Chapters 7–9: Groups, structure and patterns
Chapters 10–12: Compute, apply and diagnose
Chapter index
Sets, sums and basic bounds · Chapters 1–3
1. Begin with a set and an operation
Progressions, collisions and probability · Chapters 4–6
4. Understand why progressions have small sumsets
Groups, structure and patterns · Chapters 7–9
7. Switch to modular arithmetic without carrying the wrong bound
8. Build a higher-dimensional structured example
9. Look for three-term progressions and state the quantifiers
Compute, apply and diagnose · Chapters 10–12
10. Turn representation counts into convolution
Transfer and continue · Chapters 13–14
Let A and B be finite, nonempty sets of integers. Their sumset A + B consists of every value a + b with a in A and b in B. A sumset records possible answers, not the number of calculations that produce each answer. That distinction is the first thing to keep stable throughout the subject.
Take A = {0,2,5} and B = {0,1}. The six ordered choices produce 0,1,2,3,5,6, all different, so A + B = {0,1,2,3,5,6}. If B changes to {0,2}, the calculations give 0,2,2,4,5,7. There are still six choices, but only five distinct totals: {0,2,4,5,7}. The duplicate 2 is not discarded from the calculation history; it is counted once in the set of answers.
An additive problem must also state where addition happens. Integers have no wraparound. Residues modulo 7 do: 5 + 5 becomes 3. Vectors can be added coordinate by coordinate. Each setting has its own structural constraints, and an argument relying on ordered integers may fail in a finite group.
Sets, lists and measurements
A set forgets repeated entries. If a data file contains 2,2,5, converting it to {2,5} changes the frequencies even though it preserves which values are present. This is appropriate when asking about possible totals, but inappropriate if estimating how often totals occur under the original data distribution.
Likewise, arithmetic on measured decimal values requires a tolerance convention. Two measured sums that differ by rounding are not automatically distinct physical outcomes. The exact integer examples here avoid that issue; applying the ideas to measurements requires an explicit decision about units, precision and equality.
Before calculating, write one sentence describing the object: “A and B are sets of allowable integer durations, and repeated selections are permitted.” That sentence prevents later confusion about whether the same element can be used twice, whether order matters, and whether a total is an ordinary number or a residue.
CHAPTER 2 OF 14 · Sets, sums and basic bounds
2. Build the first complete sum table
Use A = {0,1,2,3}. For each first entry, add every possible second entry. The four rows of totals are 0,1,2,3; 1,2,3,4; 2,3,4,5; and 3,4,5,6. Thus A + A = {0,1,2,3,4,5,6}, containing seven values from sixteen ordered pairs.
For totals 0 through 6, the representation counts are respectively 1,2,3,4,3,2,1. Total 3 has four representations: (0,3), (1,2), (2,1), (3,0). Total 0 has only (0,0). This difference is invisible if the sumset alone is retained.
Now compare C = {0,1,3,7}. Its unordered pairs, allowing the same element twice, give totals 0,1,3,7,2,4,8,6,10,14. These ten totals are all distinct. Ordered pairs still number sixteen, but six off-diagonal unordered pairs each occur in two orders. C has more distinct totals than A despite having the same cardinality.
Audit both the support and the frequencies
The support is the collection of totals whose representation counts are positive. It is exactly the sumset. The sum of all representation counts must equal the number of ordered input pairs, here 16. If a computer returns ten distinct totals but only ten total representations, it has probably counted unordered pairs while labelling the output as ordered.
Neither set is intrinsically better. A process seeking many different totals may prefer C. A process seeking many interchangeable ways to obtain the same total may prefer A. The mathematical quantities clarify a tradeoff; they do not decide the practical objective.
A useful learning exercise is to reconstruct the sixteen cells before reading any theorem. Then explain why the diagonal cells have no distinct reversed partner while every off-diagonal cell does. This small observation will later explain both an upper bound for sumset size and the baseline collisions in additive energy.
CHAPTER 3 OF 14 · Sets, sums and basic bounds
3. Prove bounds before interpreting a ratio
Suppose A and B are nonempty integer sets with m and n elements. There are mn ordered choices, so |A + B| ≤ mn. For a lower bound, arrange the elements as a₁ < … < aₘ and b₁ < … < bₙ. The chain a₁+b₁ < a₂+b₁ < … < aₘ+b₁ < aₘ+b₂ < … < aₘ+bₙ contains m+n−1 distinct sums. Therefore |A + B| ≥ m+n−1.
The proof uses strict ordering and the fact that adding the same integer preserves it. It is not a valid proof in modular arithmetic, where a chain can wrap around. Naming that dependency makes the result easier to transfer correctly later.
For A = B with |A| = n, commutativity improves the upper bound. There are n diagonal choices and n(n−1)/2 unordered choices of two distinct elements. Consequently |A + A| ≤ n(n+1)/2. Combining the bounds gives 2n−1 ≤ |A + A| ≤ n(n+1)/2 for finite nonempty integer sets.
A four-element scale
For n = 4, the allowable range is 7 through 10. Our A reaches 7 and C reaches 10. This establishes that both endpoints are attainable at that size. A five-element integer set, by contrast, has at least 9 and at most 15 distinct pair sums. An output of 8 is not an unusually strong pattern; it signals a changed setting or a counting error.
The doubling constant K is |A + A|/|A|. For the four-element interval it is 7/4; for C it is 10/4. The ratio compares growth with input size, but its interpretation depends on that size. Small samples cannot have arbitrarily large doubling because the upper bound itself is limited.
To report a ratio responsibly, give the input size, sumset size and ambient number system together. “K = 2” alone is an incomplete description. It does not say whether there are ten integers, a million integers or a subgroup in a finite group.
CHAPTER 4 OF 14 · Progressions, collisions and probability
4. Understand why progressions have small sumsets
Let P = {a, a+d, …, a+(n−1)d}, where d is a nonzero integer. Adding two members gives 2a+(i+j)d with 0 ≤ i,j ≤ n−1. The index i+j takes every value from 0 through 2n−2. Thus P + P has exactly 2n−1 elements.
The overlap comes from equal index sums. For example, the third and fifth terms have the same total as the second and sixth terms when both pairs exist. Many different pairs are forced onto the same arithmetic grid. This mechanism explains the small sumset rather than merely recognising a familiar formula.
For P = {4,9,14,19}, the sumset is {8,13,18,23,28,33,38}. Translating every input by 100 translates every sum by 200, without changing the number of sums. Multiplying all inputs by a nonzero integer also preserves the number of distinct sums over the integers.
Distinguish sequence questions from set questions
A sequence has an order and may repeat values. A set of progression terms forgets their ordering and retains their values. Finding the twentieth term is a different task from counting all totals obtained by choosing two terms. Both use the same progression structure, but they ask different questions.
The existing guide to Arithmetic and Geometric Progressions develops common differences, common ratios and term formulas. Here that knowledge becomes a construction tool: the common difference forces many additive coincidences.
Do not infer an exact progression merely because a set has relatively few sums. Removing a few entries from a long progression can retain modest growth while destroying constant gaps. Higher-dimensional constructions can also have small doubling. The inverse question—what small growth forces—is subtler than recognising one example that achieves it.
CHAPTER 5 OF 14 · Progressions, collisions and probability
5. Measure collisions with additive energy
Let r(s) count the ordered pairs (a,b) in A × A satisfying a+b=s. The additive energy E(A) is the sum of r(s)² over all sums s. Equivalently, it counts ordered quadruples (a,b,c,d) in A⁴ for which a+b=c+d. Choosing two representations of the same total produces exactly such a quadruple.
For A = {0,1,2,3}, the representation counts are 1,2,3,4,3,2,1, so E(A) = 1+4+9+16+9+4+1 = 44. For C = {0,1,3,7}, four diagonal totals have one representation each and six off-diagonal totals have two each, giving E(C) = 4+6×4 = 28.
Energy therefore distinguishes the stronger concentration of pair sums in A. It is a count of arithmetic coincidences, not physical energy, and its numerical value depends strongly on the size of the input set. Comparing unnormalised energy between sets of very different sizes can be misleading.
Derive the basic inequality
There are n² ordered pairs in total, so Σr(s)=n². Cauchy–Schwarz gives (Σr(s))² ≤ |A+A| Σr(s)². Substituting yields E(A) ≥ n⁴/|A+A|. If |A+A| ≤ Kn, then E(A) ≥ n³/K.
For the interval of four points, this lower bound is 256/7, about 36.57, while the actual energy is 44. Equality in the inequality would require equal nonzero representation counts; our triangular counts are not equal. The slack is meaningful rather than a numerical mistake.
Be careful with the converse. The inequality does not say that high energy forces the whole set to have a small sumset. A structured portion can contribute many collisions while other points add numerous new sums. Stronger inverse results extract structure under carefully stated hypotheses; a one-line rearrangement cannot replace those results.
CHAPTER 6 OF 14 · Progressions, collisions and probability
6. Give energy a probability interpretation
Select a,b,c,d independently and uniformly from a finite set A of size n. There are n⁴ equally likely ordered quadruples. The probability that a+b=c+d is E(A)/n⁴. For our interval, it is 44/256 = 11/64; for C, it is 28/256 = 7/64.
These are probabilities of equal totals from two independently chosen ordered pairs. They are not the probability that a randomly chosen sumset element has several representations. That latter experiment samples totals uniformly, which is a different distribution from sampling pairs uniformly.
For A = {0,1,2,3}, pair sampling gives total 3 probability 4/16 and total 0 probability 1/16. Uniform sampling of the seven attainable totals gives both probability 1/7. A statement about “a random total” needs to identify which sampling rule is intended.
Translate the calculation into a practical question
Imagine two independent modules whose allowable integer settings are A and B, and whose combined setting is their sum. A sumset tells which combined settings can occur. Representation counts tell how many input choices lead to each setting. If each input is chosen uniformly and independently, dividing by |A||B| gives the combined-setting distribution.
If some input settings are favoured, or one module constrains the other, that division is wrong. The actual probability requires weights or joint probabilities. The exact combinatorial support can still be useful, but the frequency model has changed.
This distinction is especially valuable in simulations. An apparently uneven histogram of totals may be an expected consequence of representation multiplicities rather than a defective random generator. Before diagnosing bias, calculate the distribution implied by the sampling process. Mathematical structure can create nonuniform outcomes from perfectly uniform inputs.
CHAPTER 7 OF 14 · Groups, structure and patterns
7. Switch to modular arithmetic without carrying the wrong bound
In residues modulo q, totals are reduced to one of q classes. Let A={0,1,2,3} modulo 5. Its integer sums range from 0 to 6, but after reduction they cover {0,1,2,3,4}. Thus |A+A|=5, below the integer lower bound 7. There is no contradiction: the ambient addition has changed.
For a prime p and nonempty subsets A,B of the residues modulo p, the Cauchy–Davenport theorem gives |A+B| ≥ min(p, |A|+|B|−1). The minimum reflects the finite space available. This statement needs primality; it is not a theorem for every modulus with the same formula.
For a counterexample to indiscriminate use, take H={0,3,6} modulo 9. Adding two multiples of 3 gives another multiple of 3, and H+H=H. The sumset has size 3, whereas the incorrectly transplanted prime-modulus formula would demand at least 5. H is a subgroup, and its closure explains the failure.
Track what reduction destroys
Reducing a set modulo q can identify distinct integers. Even if the input elements remain distinct, different integer sums can collapse to the same residue. When using a modular model to study integer sums, check both kinds of collision.
For example, {0,1,3,7} has ten distinct integer sums. Modulo 7 the input 7 already merges with 0, so the model no longer has four points. Modulo 11 the inputs remain distinct, but totals 3 and 14 coincide. Neither model preserves all the original additive relations in both directions.
A computer can implement modular addition correctly while answering the wrong question. The repair is to specify the intended group in the problem statement and in the validation output. Do not leave the modulus hidden in a library function or data type.
CHAPTER 8 OF 14 · Groups, structure and patterns
8. Build a higher-dimensional structured example
A generalised arithmetic progression combines several integer directions. Consider P={i+10j : 0≤i≤2, 0≤j≤2}. Explicitly, P={0,1,2,10,11,12,20,21,22}. It has nine points arranged as three short clusters rather than one consecutive interval.
Adding two points gives i+10j with 0≤i≤4 and 0≤j≤4. These 25 values are distinct because a change of one in j shifts the value by 10, while the full i range has width only 4. Thus |P+P|=25 and its doubling constant is 25/9.
The gaps in P are not constant, yet its sumset grows by a bounded factor. The two independent index directions explain why. Each coordinate range roughly doubles; the number of combinations grows by roughly four rather than by a factor proportional to the nine input points.
Properness matters
Calling a progression “two-dimensional” does not guarantee that different coordinate choices yield different integers. If 10 is replaced by 1, many pairs (i,j) represent the same number. A proper generalised progression is one whose coordinate map is injective on the chosen box.
Our example is proper, and its doubled coordinate box remains injective as well. These are separate checks. A coordinate map can be injective on the original box but develop collisions when the box expands. Consequently, multiplying coordinate lengths is an exact count only when the relevant injectivity condition holds.
Freiman’s theorem gives an important inverse perspective for finite integer sets: bounded doubling implies containment in a generalised progression whose dimension and size relative to the original set can be bounded in terms of the doubling constant. It does not say that every small-doubling set is one ordinary progression, nor does it turn one measured ratio into a uniquely determined reconstruction.
CHAPTER 9 OF 14 · Groups, structure and patterns
9. Look for three-term progressions and state the quantifiers
A three-term arithmetic progression consists of x, x+d, x+2d. Over the integers it is nontrivial when d is nonzero. Equivalently, its endpoints and middle satisfy x+z=2y with distinct endpoints. This puts progression counting inside the same family of additive equations as sum collisions.
In {0,1,2,3}, the increasing three-term progressions are (0,1,2) and (1,2,3). In {0,1,3,7}, there are none. Check this by testing whether the midpoint of each pair of endpoints is another member. Merely seeing several small integers close together is not a proof of a progression.
If a program counts all solutions of x+z=2y, it also counts the trivial solutions x=y=z. For the four-point interval there are four trivial solutions and four ordered nontrivial ones, because each increasing progression has two endpoint orders. The total is eight, not two. All three counts are legitimate if labelled correctly.
Density is a property relative to a container
A set with 100 members is not intrinsically dense. It is dense in an interval of length 120 and sparse in an interval of length a million. Density claims must name the surrounding interval or finite group. A translation preserves an integer set’s additive relations, but an arbitrarily enlarged containing interval can make its reported density tiny.
Deep density theorems explain why sufficiently large dense sets of integers contain arithmetic patterns. Their thresholds, quantifiers and hypotheses matter. A small progression-free example does not refute an eventual theorem, and an eventual theorem does not give a convenient threshold for a small practical data set unless a usable bound is supplied.
For the connection with recurrence and dynamical systems, How Mathematics Works: Ergodic Theory offers a separate route. That connection is a research-level bridge; it should not be used to replace an elementary count that can be checked directly.
CHAPTER 10 OF 14 · Compute, apply and diagnose
10. Turn representation counts into convolution
Define the indicator of a finite integer set A to be 1 at members of A and 0 elsewhere. The convolution of the indicators of A and B at s is the sum, over integers x, of 1_A(x)1_B(s−x). Each nonzero term corresponds to a representation s=x+(s−x). Therefore convolution computes the ordered representation counts.
This is a change of language, not a new counting convention. The support of the convolution is A+B. The sum of its coefficients is |A||B|. For A=B, the sum of squared coefficients is additive energy. Keeping those three operations distinct prevents support size, total mass and concentration from being confused.
For nonnegative integers, the same computation appears in polynomial multiplication. If A={0,1,3}, write f(t)=1+t+t³. Squaring gives 1+2t+t²+2t³+2t⁴+t⁶. There are six nonzero coefficients, nine total coefficient units and squared-coefficient sum 15. These correspond respectively to six attainable sums, nine ordered pairs and energy 15.
Choose an algorithm that matches the setting
For small sets, two nested loops and a dictionary of counts are easy to audit. For large sets in a short integer range, array convolution may be more efficient. Sparse sets with enormous gaps may make a dense array wasteful even when the number of points is modest.
A circular convolution computes sums modulo the array length. An ordinary integer sum computation requires enough padding to prevent wraparound. For inputs supported from 0 to M, pair sums can reach 2M, so the output needs positions through 2M. An undersized circular array silently merges totals.
Floating-point transforms can introduce rounding error when used to recover integer counts. Validate that recovered coefficients are close enough to integers under a justified error bound, or use exact methods appropriate to the problem. Faster output is not a substitute for knowing which arithmetic the algorithm performed.
CHAPTER 11 OF 14 · Compute, apply and diagnose
11. Work through a two-stage duration problem
Suppose an illustrative workshop offers a first activity lasting 10,20 or 30 minutes and a second lasting 5,15 or 25 minutes. Assume one activity of each kind is selected, the durations add exactly, and there is no changeover time. The possible totals are 15,25,35,45 and 55 minutes.
There are nine ordered selections. Their representation counts, in increasing total order, are 1,2,3,2,1. The 35-minute option has three implementations: (10,25), (20,15) and (30,5). A planner looking for alternative ways to fill the same duration should retain those representations rather than only the five totals.
If the second-stage durations change to 6,17 and 29 minutes, the nine totals become 16,27,39,26,37,49,36,47,59. They are all different. The revised menu offers more distinct totals but removes redundancy. Which menu is preferable depends on whether breadth of choice or interchangeability is the main objective.
Add a constraint and rebuild the object
Now suppose the 30-minute first activity is incompatible with the 5-minute second activity. The original Cartesian product of all nine pairs is no longer the allowed input space. Total 35 still occurs twice, but its third implementation disappears. The attainable-total set remains unchanged even though operational flexibility decreases.
This example shows why a sumset alone can miss a material change. If instead every implementation of one total were forbidden, the support would change too. A constrained pair relation must be represented explicitly; simply retaining A and B cannot encode which cross-pairs are prohibited.
Finally, add a fixed five-minute changeover. Every total shifts by five and representation counts are unchanged. If changeover time depends on the chosen pair, the transformation is no longer a uniform translation and must be recalculated. Identify the mechanism of the change before reusing an old result.
CHAPTER 12 OF 14 · Compute, apply and diagnose
12. Diagnose misleading conclusions and repair the evidence
The most common error is equating many calculations with many distinct results. Repair it by retaining a count dictionary and reporting both its number of keys and the sum of its values. Another frequent error is mixing ordered and unordered choices. Repair it by explaining the diagonal and reverse-pair conventions before counting.
A third error is applying an integer bound modulo a composite number. The subgroup example modulo 9 is a compact diagnostic. A fourth is treating a small doubling constant as proof of an exact ordinary progression. The nine-point, two-direction example supplies a direct counterexample to that interpretation.
A set can contain structure without being all structure
Consider a long consecutive interval together with several widely separated extra points. Pairs inside the interval still contribute many equal-sum quadruples. Pairs involving the extra points can create many new totals. The resulting set may have a prominent structured component without behaving globally like one compact progression.
To investigate such a case, separate the contributions from interval–interval, interval–outlier and outlier–outlier pairs. Do not remove inconvenient points merely to improve a statistic. If a subset is studied, report how it was chosen and what fraction of the original data it retains.
This is the same intellectual discipline developed in How to Check a Number Pattern When Several Rules Fit the Same Terms. A pattern that fits observed values is a candidate explanation, while a theorem or specified generating rule supports a stronger claim.
Distinguish audit failures from mathematical surprises
When a result violates a proved bound, first inspect the definition and implementation. Check duplicates in the input, restricted versus unrestricted pairs, modular reduction, floating-point equality and accidental omission of diagonal pairs. Only after those checks should a genuinely different mathematical setting be considered.
For reproducible work, preserve the exact input sets, arithmetic convention, representation histogram and verification script. A single summary ratio cannot reveal where an error occurred. The full small certificate often teaches more than a large unexplained plot.
CHAPTER 13 OF 14 · Transfer and continue
13. Test transfer with changed-context questions
Try each question before reading its answer. The point is to decide which feature survives the change, not merely to insert new numbers into the last formula used.
Translation and scaling
Let A={0,1,2,3} and D={10,13,16,19}. How many sums does D+D contain, and what is its energy? Answer: D=10+3A. Every sum transforms as s↦20+3s, an injective transformation over the integers. Therefore there are seven sums and energy 44. The representations are relabelled, not changed.
Removing repetition of the same entry
For A={0,1,2,3}, allow only pairs with distinct entries. What are the possible totals? Answer: {1,2,3,4,5}, with ordered representation counts 2,2,4,2,2. There are twelve ordered pairs. The squared-count sum is 32, but it describes this restricted relation rather than the unrestricted additive energy defined earlier.
Changing the ambient group
For A={0,1,2,3} modulo 5, find the representation counts for residues 0 through 4. Answer: 3,3,3,4,3. The integer counts for totals 5 and 6 are folded into residues 0 and 1. Their sum is 16, and their squared sum is 52. Modular collisions increase the count of equal-sum quadruples in this example.
Testing a new polynomial
Let B={0,2,4}. What does (1+t²+t⁴)² tell you? Answer: it is 1+2t²+3t⁴+2t⁶+t⁸. There are five attainable sums, nine ordered pairs and energy 19. Sampling pairs uniformly gives total 4 probability 3/9; sampling attainable totals uniformly gives probability 1/5.
Explain why your method still applies
A complete response names the changed assumption. Translation and nonzero integer scaling preserve equal-sum relations. Removing diagonal pairs changes the permitted input relation. Modular reduction changes equality of totals. Changing the sampling scheme changes probabilities even when attainable values remain the same.
These are different operations with different consequences. Being able to distinguish them is more useful than memorising one answer from the original four-point example. It also provides a practical checklist for reviewing another person’s calculation without relying on their software.
CHAPTER 14 OF 14 · Transfer and continue
14. Learning routes, common questions and primary reading
Is additive combinatorics just adding sequences?
No. Sequence formulas are an accessible entry point, but the subject studies sets, additive equations, growth, patterns and structural implications. It connects combinatorial counting with number theory, algebra and analysis. The guide to Additive versus Multiplicative Number Patterns helps establish the operation before moving into these set questions.
Does A+A mean doubling every element?
No. A+A allows two independently chosen members. The dilation {2a:a∈A} uses the same member twice. For A={0,1,2}, the sumset is {0,1,2,3,4}, while the dilation is {0,2,4}. Some authors use 2A for the iterated sumset, so check notation rather than assuming it means dilation.
Does a large sumset mean randomness?
Not by itself. A deliberately constructed set can have many distinct pair sums. Randomness concerns a probability model, while sumset size is a deterministic property of the supplied set. A claim that a set resembles a random sample needs a stated comparison model and more evidence than one statistic.
Where should a beginner start?
First construct sum tables for three- and four-element sets. Then derive the integer lower bound using an increasing chain and the upper bound using unordered pairs. Next compute representation histograms and energy. Only after those quantities feel concrete should inverse theorems and higher-dimensional progressions be introduced.
For an algebra route, compare polynomial multiplication with the count dictionary. For a computational route, implement both methods and test their agreement on small sets, including negative values and modular examples where appropriate. For an applications route, choose a real pair-selection problem and distinguish possible totals, frequencies and forbidden combinations.
The adjacent Design Theory root asks how combinations can be distributed with prescribed balance. The connection is the discipline of counting relationships explicitly: additive combinatorics groups pairs by equal totals, while block-design problems group objects by membership. Their definitions differ, but both reward independent audits of what was actually counted.
Primary sources and further study
Yufei Zhao’s MIT notes, Structure of Set Addition, develop sumset inequalities, generalised progressions, Freiman’s theorem and additive energy. They provide the technical route beyond the explicit elementary calculations here. Historical notes about open problems should be read in their publication context rather than treated as a current research-status list.
John Griesmer’s Structure in Sumsets seminar overview, hosted by Colorado State University, states the prime-modulus Cauchy–Davenport theorem and describes direct and inverse questions. It is a useful orientation to the difference between predicting a sumset from its inputs and inferring input structure from a sumset.
A successful learner can finish with a compact certificate: the exact sets, the addition rule, the attainable totals, the representation counts and a justified conclusion. When the context changes, rebuild only the parts whose assumptions changed—but check those parts fully. That is how a subject about addition becomes a reliable method of reasoning about structure.