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Proof step 8 from proof case-10 (previous step)

1 < i, 0 < i, 2 < i, i < 8 ⊦ 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 9:
⊦ i < j ↔ i +1 = j ∨ i +1 < j
Proof step 10:
1 < i, 0 < i, 2 < i, i +1 = 8 ∨ i +1 < 8 ⊦ i = 9, i = 8, i = 7, i = 6, i = 5, i = 4, i = 3, i = 2

Rule argument

spec: int-basic1 inst: name: lsrec seq: ⊦ i < j ↔ i +1 = j ∨ i +1 < j subst: [j, i] → [8, i] flags: rewrite

Additional information

Used terms: 8, i