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Proof step 14 from proof lst-02 (previous step)

j0 = i +1, 0 < j0 ∨ j0 < 0 ⊦ 0 = j0, 1 = j0, i < 0, 0 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < i +1 ∨ i +1 < 0 ⊦ i = 0, i = ~1, 0 < i, i < 0

Rule argument

Additional information

Used simplifier rules: numeral: ⊦ i +1 = (* j) ↔ i = j -1;