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Proof step 19 from proof gcdex (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: ~ i0, ~ j0 abort: F discard: T computed: F

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-11: ⊦ ~ i + ~ j = ~(i + j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs Used terms: ~ i0, ~ j0