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Proof step 23 from proof three (previous step)

i * j = 3, j < 3, 0 < i, 1 < i, 1 < j, i < 3, 0 < j ⊦ i = 3, j = 3, j = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ i < j ↔ i +1 = j ∨ i +1 < j

Rule argument

spec: int-basic1 name: lsrec seq: ⊦ i < j ↔ i +1 = j ∨ i +1 < j subst: [j, i] → [3, j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ 0 < j → (j < 2 ↔ j = 1); (s) Used terms: 3, j