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Proof step 8 from proof ple-01 (previous step)

i0 < j0, i1 + 1 < j ⊦ (i1 + 1) + i0 < j + j0, i1 < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)
Proof step 10:
∃ i1. j0 = i0 + i1 ∧ 0 < i1, i1 + 1 < j ⊦ (i1 + 1) + i0 < j + j0, i1 < j

Rule argument

spec: int-basic2 inst: name: ex-ls-add seq: ⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0) subst: [j, i] → [j0, i0] flags: rewrite

Additional information

Used terms: j0, i0