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Proof step 12 from proof gcdhex (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: 0, 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: 0, 1