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Proof step 24 from proof case-20 (previous step)

1 < i, 0 < i, 10 < i, i < 13 ⊦ i = 10, i = 11, i = 12, i = 13, i = 14, i = 15, i = 16, i = 17, i = 18, i = 19

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ i ≠ j -1 → (i < j ↔ i < j -1)
Proof step 26:
i < 13 -1, 10 < i, 0 < i, 1 < i ⊦ i = 19, i = 18, i = 17, i = 16, i = 15, i = 14, i = 13, i = 12, i = 11, i = 10, i = 13 -1

Rule argument

spec: int-basic1 name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [13, i] flags: <all paths>

Additional information

Used terms: 13, i