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Proof step 45 from proof equal-minus-one (previous step)

0 < ~ j, i ^ (~ j)-1 * i = ~1, j < 0 ⊦ i = 0, i = ~1, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
abs(i ^ ~(j + 1)) = 1, i + i ^ ~(j + 1) = 0, j < 0 ⊦ i = 0, i = ~1, j = 0

Rule argument

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%eq-one-neg: ⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j; from specification int-mult%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%sub-pred: ⊦ i -1 = i - 1; from specification int-abs