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Proof step 14 from proof gcd-mod-one (previous step)

i * i0 = j * i1 + 1, gcd(i, j) = 1, i1 < 0, 1 < i, 0 < i, 1 < j, 0 < j, j * i1 < 0, i * i0 ≤ 0, 0 ≤ i0 ⊦ (i * i0) % j = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: gcd-one-01: ⊦ gcd(i, 1) = 1;%j ≠ 0 ⊦ 0 % j = 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ 1 < i → abs(i) = i; (s)% ⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j; (ws)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ i * 0 = 0; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0); (s)