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Proof step 11 from proof prime-5 (previous step)

5 % i = 0, 1 < i, 0 < i, i < 3, i * i ≤ 5 ⊦ i = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ i ≠ j -1 → (i < j ↔ i < j -1)

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: ⊦ 0 < j → (j < 2 ↔ j = 1); (s)% ⊦ i = 1 → j % i = 0; (s)% ⊦ i = 1 → i * j = j; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs Used terms: 3, i