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Proof step 23 from proof gcdex (previous step)

gcd(abs(i), abs(j)) = abs(i) * i0 + abs(j) * j0, abs(j) ≤ abs(i) ⊦ j < 0, i < 0, ∃ i0, j0. gcd(abs(i), abs(j)) = i * i0 + j * j0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: i0, j0 abort: F discard: T computed: F

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs Used terms: i0, j0