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Proof step 3 from proof abs-pred-01 (previous step)

i→n(i) = m + 1 ∧ i→n(i) ≠ 0, i < 0 ⊦ i = ~1, i→n(i + 1) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;