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Proof step 13 from proof one-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs