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The proof that S(k) is true for all k ≥ 12 can then be achieved by induction on k as follows: Base case: Showing that S(k) holds for k = 12 is simple: take three 4-dollar coins. Induction step: Given that S(k) holds for some value of k ≥ 12 (induction hypothesis), prove that S(k + 1) holds, too. Assume S(k) is true for some arbitrary k ≥ 12.
P. Oxy. 29, one of the oldest surviving fragments of Euclid's Elements, a textbook used for millennia to teach proof-writing techniques. The diagram accompanies Book II, Proposition 5. [1] A mathematical proof is a deductive argument for a mathematical statement, showing that the stated assumptions logically guarantee the
A step-by-step proof. All Metamath proof steps use a single substitution rule, which is just the simple replacement of a variable with an expression and not the proper substitution described in works on predicate calculus. Proper substitution, in Metamath databases that support it, is a derived construct instead of one built into the Metamath ...
We prove associativity by first fixing natural numbers a and b and applying induction on the natural number c.. For the base case c = 0, (a + b) + 0 = a + b = a + (b + 0)Each equation follows by definition [A1]; the first with a + b, the second with b.
Fermat's little theorem and some proofs; Gödel's completeness theorem and its original proof; Mathematical induction and a proof; Proof that 0.999... equals 1; Proof that 22/7 exceeds π; Proof that e is irrational; Proof that π is irrational; Proof that the sum of the reciprocals of the primes diverges
For each step S in the deduction that is a premise in Γ (a reiteration step) or an axiom, we can apply modus ponens to the axiom 1, S→(H→S), to get H→S. If the step is H itself (a hypothesis step), we apply the theorem schema to get H→H.