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Proof step 5 from proof lst-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j0 ∨ j0 < 0 ⊦ 0 = j0, j0 = ~1, 0 < j0 +1, j0 +1 < 0

Rule argument

Additional information

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