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Proof step 11 from proof gcdhex (previous step)

i = j, gcd(i, j) = gcd(j, i % j), 0 ≤ i, j ≤ i, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < j, ∀ i0, j0. i→n(i0) < i→n(j) → 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 ≤ i0 → (∃ i1, j1. gcd(i0, j0) = i0 * i1 + j0 * j1) ⊦ j = 0, ∃ i, j0. j = j * i + j * j0

Rule argument

Additional information

Used simplifier rules: gcd-same-01: ⊦ 0 < j → gcd(j, j) = j;%gcd-zero-03: ⊦ i ≠ 0 → gcd(i, 0) = abs(i);%ler: ⊦ i ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%mod-same: i ≠ 0 ⊦ i % i = 0; from specification int-div