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Proof step 3 from proof rax-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 < 0 ⊦ i→n(~ i0) = i→n(i0 + 1) + 1

Rule argument

Additional information

Used simplifier rules: Int2nat-neg: ⊦ i < 0 → i→n(i) = i→n(~ i);%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2