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Proof step 41 from proof gcd-rec-sub-03 (previous step)

i < 0 ⊦ 0 < abs(i) - abs(i0) ∧ abs(i) - abs(i0) < abs(i), abs(i) < i0, i0 < 0, gcd(abs(i) - abs(i0), abs(i)) = gcd(abs(i) - abs(i0), abs(i0)), 0 < i ∧ i < i0 + i

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs