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Proof step 3 from proof case-30 (previous step)

1 < i, 0 < i, 20 < i, i < 30 ⊦ i = 20, i = 21, i = 22, i = 23, i = 24, i = 25, i = 26, i = 27, i = 28, i = 29

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 < 30 -1, 20 < i, 0 < i, 1 < i ⊦ i = 29, i = 28, i = 27, i = 26, i = 25, i = 24, i = 23, i = 22, i = 21, i = 20, i = 30 -1

Rule argument

spec: int-basic1 name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [30, i] flags: <all paths>

Additional information

Used terms: 30, i