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Proof step 6 from proof rax-03 (previous step)

0 < ~ i0, i0 < 0 ⊦ i→n(~ i0 - 1) + 1 = i→n(i0 + 1) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i0 < 0 ⊦ i→n(~ i0 - 1) = i→n(i0 + 1)

Rule argument

Additional information

Used simplifier rules: neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2