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Proof step 5 from proof gcdhex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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) ⊦ ∃ i0. i = i * i0

Rule argument

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * 0 = 0; (s)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)