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Proof step 42 from proof equal-minus-one (previous step)

0 < ~(i ^ ~ j) ∧ ~(i ^ ~ j) ≤ 1, i ^ ~ j < 0, ~((1 - ~(i ^ ~ j)) / ~(i ^ ~ j) + 1) = ~1, j < 0, i ^ ~ j < 0 ⊦ j = 0, i = ~1, i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero-02: ⊦ j < 0 → (i ^ ~ j = 0 ↔ i = 0);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%minus-sub: ⊦ ~(i - j) = j - i; from specification int-abs%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%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j); from specification int-div%subneg: ⊦ i - ~ j = i + j; from specification int-abs