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Proof step 5 from proof div-one-one-neg (previous step)

∃ i0. i * i0 = ~1 ⊦ i = 1, i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs(i0) = 1, i0 + i = 0 ⊦ i = 1, i = ~1

Rule argument

Additional information

Used simplifier rules: 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%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%eq-one-neg: ⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j; from specification int-mult