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Proof step 47 from proof lagrange-lemma (previous step)

j * i = (j1 ^2 + j0 ^2) + 1, j1 ^2 ≤ (j - 1) / 2 * (j - 1) / 2, j0 ^2 ≤ (j - 1) / 2 * (j - 1) / 2, 0 ≤ j1, 0 ≤ j, prime(j), j0 ≤ (j - 1) / 2, j1 ≤ (j - 1) / 2, 0 ≤ j0, 0 < j, 0 < i, 0 ≤ i, j * j ≤ (2 * (j - 1) / 2) * (j - 1) / 2 + 1, j ≤ i, j * i ≤ (2 * (j - 1) / 2) * (j - 1) / 2 + 1 ⊦ j = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 48:
j ≠ 0 ⊦ i % j = 0 → i / j * j = i
Proof step 49:
2 ≠ 0 ∧ (j - 1) % 2 = 0, j * i = (j1 ^2 + j0 ^2) + 1, j1 ^2 ≤ (j - 1) / 2 * (j - 1) / 2, j0 ^2 ≤ (j - 1) / 2 * (j - 1) / 2, 0 ≤ j1, 0 ≤ j, prime(j), j0 ≤ (j - 1) / 2, j1 ≤ (j - 1) / 2, 0 ≤ j0, 0 < j, 0 < i, 0 ≤ i, j * j ≤ (j - 1) * (j - 1) / 2 + 1, j ≤ i, j * i ≤ (j - 1) * (j - 1) / 2 + 1 ⊦ j = 2

Rule argument

spec: int-div inst: name: divmult seq: j ≠ 0 ⊦ i % j = 0 → i / j * j = i subst: [j, i] → [2, j - 1] flags: rewrite

Additional information

Used simplifier rules: prime-odd-01: ⊦ prime(j) ∧ j ≠ 2 → (j - 1) % 2 = 0; from specification int-gcd%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2, j - 1