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Proof step 11 from proof case-10 (previous step)

1 < i, 0 < i, 2 < i, i < 7 ⊦ i = 9, i = 8, i = 7, i = 6, i = 5, i = 4, i = 3, i = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ i ≠ j -1 → (i < j ↔ i < j -1)
Proof step 13:
i ≠ 7 -1, 1 < i, 0 < i, 2 < i, i < 7 -1 ⊦ i = 9, i = 8, i = 7, i = 6, i = 5, i = 4, i = 3, i = 2

Rule argument

spec: int-basic1 inst: name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [7, i] flags: rewrite

Additional information

Used terms: 7, i