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Proof step 14 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) ⊦ 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 15:
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
Proof step 18:
∃ i1, j0. gcd(j, i % j) = j * i1 + i % j * j0, 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) ⊦ 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: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j, i % j