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Proof step 14 from proof one-neg (previous step)

1 = i0 * j + j0, j < 0, i0 < j0, j < j0, j0 + j < 0, i0 < 0, i0 + j < ~1, 0 ≤ j0 ⊦ j = 0, j = ~1, i0 * ~ j + 1 = 0 * ~ j + j0 ∧ 0 ≤ 1 ∧ 1 < ~ j ∧ j0 < ~ j

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + 0 = i; (s)% ⊦ i * 0 = 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; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ ~ i * j = ~(i * j); (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)