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Proof step 15 from proof gcdhex (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 16:
gcd(i, j) = gcd(j, i % j), 0 < i, 1 < i, j < i, 0 < j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → j ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ i = j, j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0, i % j ≤ j ∧ 0 ≤ i % j ∧ 0 ≤ j
Proof step 19:
gcd(i, j) = gcd(j, i % j), 0 < i, 1 < i, j < i, 0 < j, 0 ≤ i, ∃ i1, j0. gcd(j, i % j) = j * i1 + i % j * j0, ∀ i0, j. i→n(i0) < i→n(i) → j ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ 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

subst: [i0, j] → [j, i % j]

Additional information

Used simplifier rules: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: j, i % j