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Proof step 9 from proof gcdex-smallest-pos (previous step)

gcd(i, j) = i * i0 + j * j0, gcd(i, j) = i * i1 + j * j1, 0 ≤ i0, i1 < i0, 0 ≤ i1, 0 < i0, ∀ i1, i, j, j0. i→n(i1) < i→n(i0) → gcd(i, j) = i * i1 + j * j0 ∧ 0 ≤ i1 → j = 0 ∨ (∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 ∧ (∀ i1. 0 ≤ i1 ∧ i1 < i0 → ¬(∃ j1. gcd(i, j) = i * i1 + j * j1))) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 ∧ (∀ i1. 0 ≤ i1 ∧ i1 < i0 → ¬(∃ j1. gcd(i, j) = i * i1 + j * j1))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i1, i, j, j0] → [i1, i, j, j1]

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%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%absle: ⊦ 0 ≤ i → abs(i) = i; 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: i1, i, j, j1