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Proof step 4 from proof abs-zero (previous step)

¬ 0 < i, i→n(i) = 0 ⊦ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i→n(~ i) = 0, i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: Int2nat-neg: ⊦ i < 0 → i→n(i) = i→n(~ i);%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1