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Proof step 15 from proof gcdhex (previous step)

gcd(i, j) = gcd(j, i % j), 0 < j, 0 ≤ i, j < i, 1 < i, 0 < i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ 0 ≤ j ∧ 0 ≤ i % j ∧ i % j ≤ j, i = j, j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
gcd(i, j) = gcd(j, i % j), 0 < j, 0 ≤ i, j < i, 1 < i, 0 < i, j < i % j, 0 < i % j, 1 < i % j, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ i = j, j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Rule argument

Additional information

Used simplifier rules: trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%mod-zero: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j; from specification int-div%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs