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Proof step 9 from proof square-le-one (previous step)

0 < i, 0 < sqrt(i), 1 < sqrt(i), i < 3 ⊦ i = 3

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i ≠ j -1 → (i < j ↔ i < j -1)
Proof step 11:
i = 3 -1, i < 3, 1 < sqrt(i), 0 < sqrt(i), 0 < i ⊦ i = 3

Rule argument

spec: int-basic1 name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [3, i] flags: <all paths>

Additional information

Used simplifier rules: one-01: ⊦ sqrt(1) = 1;% ⊦ 0 < j → (j < 2 ↔ j = 1); (s) Used terms: 3, i