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Proof step 11 from proof lst-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
⊦ i = j -1 ↔ j = i +1
Proof step 13:
j0 = i +1, j0 < 0 ∨ 0 < j0 ⊦ 1 = j0, 0 = j0, i < 0, 0 < i

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [j0, i] flags: rewrite, current

Additional information

Used terms: j0, i