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Proof step 3 from proof case-10 (previous step)

1 < i, 0 < i, 2 < i, i < 10 ⊦ i = 2, i = 3, i = 4, i = 5, i = 6, i = 7, i = 8, i = 9

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ≠ j -1 → (i < j ↔ i < j -1)
Proof step 5:
i < 10 -1, 2 < i, 0 < i, 1 < i ⊦ i = 9, i = 8, i = 7, i = 6, i = 5, i = 4, i = 3, i = 2, i = 10 -1

Rule argument

spec: int-basic1 name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [10, i] flags: <all paths>

Additional information

Used terms: 10, i