From Individual Lemmas to a Proof Plan
In Writing Lemmas and Corollaries, a lemma was treated as a precise, reusable result that can support a larger argument. A proof with several difficult steps can use several such results. The important organizational question is how those results depend on one another: which conclusion becomes a hypothesis or useful fact in the next step?
A chain of lemmas is a sequence of results arranged so that one established step enables another. The first lemma uses the original assumptions. Its conclusion supplies information for the second lemma, whose conclusion supplies information for the third, and so on, until the desired claim is reached. The final proof is valid only if every link is valid and the links actually connect.
This organization is useful even when the finished proof is short. Before writing, one can identify the target, ask what would be enough to prove it, and continue backward until reaching facts that follow directly from the hypotheses. Then the written argument can present those steps in the forward order in which they are justified.
The Logical Form of a Chain
The basic logical pattern can be written without any particular mathematical subject. Suppose \(P\), \(Q\), and \(R\) are propositions, and suppose \(P\) implies \(Q\), while \(Q\) implies \(R\). Then \(P\) implies \(R\). The middle proposition is the link: it is obtained from the first claim and is exactly what the next implication needs.
For a longer chain, the same reasoning is repeated. If \(P_0\) implies \(P_1\), \(P_1\) implies \(P_2\), and each subsequent proposition implies the next, then \(P_0\) implies the final proposition. This follows by repeated use of the transitivity of implication. The chain does not permit a missing link: knowing \(P_0\to P_1\) and \(P_2\to P_3\) does not establish \(P_0\to P_3\) unless there is a justified route from \(P_1\) to \(P_2\).
Here is the general statement and its proof.
Proposition (Chaining Implications). Let \(P_0,P_1,\ldots,P_n\) be propositions, where \(n\geq1\). If \(P_i\to P_{i+1}\) holds for every integer \(i\) with \(0\leq i<n\), then \(P_0\to P_n\).
Proof. We use induction on \(n\). If \(n=1\), the conclusion \(P_0\to P_1\) is exactly the assumed implication for \(i=0\). Now suppose the result holds for a chain of \(n\) implications. Consider a chain of \(n+1\) implications. The first \(n\) implications give \(P_0\to P_n\) by the induction hypothesis. The final assumed implication is \(P_n\to P_{n+1}\). By transitivity of implication, \(P_0\to P_{n+1}\). This proves the claim for \(n+1\), and therefore for every \(n\geq1\).
In a mathematical proof, the propositions are often statements about numbers or objects, and the links are lemmas. The same logical discipline applies. Record the assumptions available at each stage; do not use a later conclusion before it has been established. A lemma may also need hypotheses that do not appear in the preceding lemma's conclusion. Such hypotheses must come from the original assumptions or from another proved result.
Design Each Link Around Its Input and Output
A useful planning method is to write a short list of the states of knowledge the proof must pass through. Each state should be precise enough to check. For example, if the final goal requires multiplying an inequality by a positive number, a helpful intermediate result may establish that the multiplier is positive. Merely knowing that an inequality holds is not enough if the next operation also needs a sign condition.
The conclusion of a lemma should expose the information needed later, not just a fragment of it. If the next step needs both \(u<v\) and \(r>0\), then the chain must preserve or re-establish both facts. One can keep the original assumptions in force throughout a proof, but the written argument should make clear which facts are being carried forward.
Write the exact statement you want to prove, including its domain and all boundary conditions.
Identify a nearby fact that would imply the endpoint, and note any conditions that fact requires.
Continue identifying needed facts until each one follows directly from the assumptions or an established result.
Establish the first available facts, then use each result only after its hypotheses have been verified.
For each adjacent pair of steps, confirm that the earlier conclusion really supplies what the later argument needs.
A backward plan can help discover lemmas, but the finished proof should not leave the reader to fill in the order of justification. State or cite the result that supports each step, and make any substitution explicit. The goal is not to insert a separate lemma heading for every sentence. A lemma is useful when it isolates a meaningful reusable step or makes a dependency clearer.
Worked Example: A Chain of Strict Inequalities
We first establish a reusable fact about multiplying inequalities.
Lemma. Let \(u,v,r\in\mathbb R\). If \(u<v\) and \(r>0\), then \(ru<rv\).
Proof. Since \(u<v\), we have \(v-u>0\). Since \(r>0\), the product of the positive real numbers \(r\) and \(v-u\) is positive: $$ r(v-u)>0. $$ Expanding gives \(rv-ru>0\), which is equivalent to \(ru<rv\). This proves the lemma.
Now let \(x\in\mathbb R\) and suppose \(0<x<1\). We will prove the chain \(x^3<x^2<x\). First, \(x>0\) and \(x<1\). Apply the lemma with \(u=x\), \(v=1\), and \(r=x\). The hypotheses hold, so $$ x^2<x. $$ Also, \(x^2>0\), because \(x>0\) and the product of two positive real numbers is positive. Apply the lemma again with \(u=x\), \(v=1\), and \(r=x^2\). This gives $$ x^3<x^2. $$ Together these two inequalities give \(x^3<x^2<x\). Each application requires a positive multiplier, and the proof establishes that condition before using it.
When a Proof Branches Before It Joins
Not every proof has the shape of one straight line. A later step may need two separate facts, each proved from the same original assumptions. The structure then branches: one line establishes the first fact, another establishes the second, and a later result combines them. This is still a dependency structure, but it is not a single chain of propositions.
For example, a proof might use one lemma to show \(A\), another to show \(B\), and then a third result whose hypotheses are \(A\) and \(B\). To apply the third result, both branches must be complete. Showing only \(A\) does not justify a conclusion whose hypotheses require \(A\) and \(B\). Conversely, if one branch already gives all the information needed, proving an unnecessary second branch may make the argument longer without making it stronger.
Worked Example: Comparing Two Positive Products
Let \(a,b,c,d\in\mathbb R\), and suppose \(0<a<b\) and \(0<c<d\). We will prove \(ac<bd\) by inserting an intermediate quantity.
First, apply the multiplication lemma with \(u=a\), \(v=b\), and \(r=c\). Its hypotheses hold because \(a<b\) and \(c>0\). Therefore, $$ ac<bc. $$ For the next step, use \(c<d\) and the positive multiplier \(b\), which is positive because \(0<a<b\). The lemma with \(u=c\), \(v=d\), and \(r=b\) gives $$ bc<bd. $$ By transitivity of the real-number order, \(ac<bd\). The inserted term \(bc\) is the link between the two applications: it is the conclusion of the first comparison and the starting expression in the second.
This proof also shows why the two positivity assumptions matter. The first multiplication uses \(c>0\), while the second uses \(b>0\). If a multiplier were negative, multiplying an inequality by it would reverse the order; if it were zero, the strict inequality would become equality. The hypotheses are checked where they are needed rather than treated as decorative conditions.
Worked Example: Two Branches That Meet at a Sum
Suppose \(x\in\mathbb R\) and \(0<x<2\). We will prove \(x^2+x<6\). This argument has two branches that meet when the inequalities are added.
For the first branch, apply the multiplication lemma to \(x<2\) using the positive multiplier \(x\). Since \(x>0\), it gives $$ x^2<2x. $$ For the second branch, the original hypothesis already gives \(x<2\). These two inequalities alone do not yield the desired bound by adding, so we use a more suitable first branch: apply the multiplication lemma to \(x<2\) with multiplier \(2\), obtaining \(2x<4\). Since \(x<2\), adding these strict inequalities gives $$ 2x+x<4+2, $$ so \(3x<6\). This does not itself establish the stated target \(x^2+x<6\), so we instead use the original upper bound in a product comparison.
Because \(0<x<2\), we have \(0<x<2\) and \(0<x+1<3\). Apply the product-comparison argument from the preceding example with \(a=x\), \(b=2\), \(c=x+1\), and \(d=3\). The required inequalities hold, so $$ x(x+1)<2\cdot3=6. $$ Since \(x(x+1)=x^2+x\), this proves \(x^2+x<6\).
The proof illustrates why checking the endpoint of a planned chain matters. The first attempted pair of estimates above gives \(3x<6\), not \(x^2+x<6\); that endpoint is insufficient. The product comparison supplies the correct connection and reaches the exact target. In a finished proof, one would omit the unsuccessful exploratory route and present the justified product comparison directly.
Check Every Dependency Before You Finish
A chain can fail even when each individual sentence looks plausible. The most common problem is a mismatch: a lemma establishes a statement similar to the next step's hypothesis, but not the same statement. Another is an unstated condition, such as positivity before multiplication. A third is an order error, where the conclusion of a later step is used before it has been proved.
A final audit should follow the argument one link at a time. For each lemma, identify its inputs and verify every hypothesis. Then compare its conclusion with the next step: is it exactly what is needed, or is a short additional deduction required? At the end, compare the final conclusion with the original goal, including whether the inequality is strict or weak and whether all variables are in the stated domain.
The chain should also be efficient. If two adjacent statements can be combined without obscuring the reasoning, a separate lemma may not be necessary. On the other hand, keeping an intermediate result can be valuable when it is independently reusable, when it isolates a delicate hypothesis, or when a longer proof becomes easier to verify in smaller parts. The aim is a visible route from assumptions to conclusion, not the largest possible number of labels.
Check Your Understanding
For each question, consider both the logical connection and the hypotheses needed at each link.
- If \(P\to Q\) and \(Q\to R\), what proposition serves as the link between the first implication and the second?
- In the proof that \(x^3<x^2<x\) for \(0<x<1\), why must the positivity of \(x^2\) be established before the second application of the multiplication lemma?
- In the comparison \(ac<bc<bd\), identify which multiplier is used in each application and which hypothesis makes it positive.
- Why did the intermediate estimate \(3x<6\) fail to prove \(x^2+x<6\) on its own?
- A lemma concludes \(A\), while the next result requires both \(A\) and \(B\). What additional step must the proof contain before that result can be applied?
- Give one reason to retain an intermediate lemma and one reason that a separate lemma heading might be unnecessary.