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Proof step 6 from proof int-simp-04 (previous step)

i2 < i3, i0 + i3 < i2, 0 < i0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)
Proof step 8:
0 < i0, i0 + i3 < i2, ∃ i0. i3 = i2 + i0 ∧ 0 < i0 ⊦

Rule argument

spec: int-basic2 name: ex-ls-add seq: ⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0) subst: [j, i] → [i3, i2] flags: <all paths>

Additional information

Used terms: i3, i2