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Proof step 5 from proof int-simp-06 (previous step)

i0 + i3 < i2, 0 < i0 ⊦ i3 < i2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)
Proof step 7:
i0 + i3 < i2, 0 < i0 ⊦ ∃ i0. i2 = i3 + i0 ∧ 0 < i0

Rule argument

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

Additional information

Used terms: i2, i3