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Proof step 5 from proof minus-one (previous step)

~2 < i, i < ~1, i < 0 ⊦ i = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < j ↔ i +1 = j ∨ i +1 < j

Rule argument

spec: int-basic1 inst: name: lsrec seq: ⊦ i < j ↔ i +1 = j ∨ i +1 < j subst: [j, i] → [~1, i] flags: rewrite, current

Additional information

Used simplifier rules: numeral: ⊦ i +1 = (* j) ↔ i = j -1;%numeral-01: ⊦ i +1 < (* j) ↔ i < j -1;%ls-01: ⊦ i < j → ¬ j < i; Used terms: ~1, i