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Proof step 7 from proof rax-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i0 < 0 ⊦ i→n(~ i0 - 1) = i→n(i0 + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)