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Proof step 3 from proof int-simp-08 (previous step)

i2 < i3, i < i1 ⊦ i + i2 < i1 + i3

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)
Proof step 5:
i2 < i3, ∃ i0. i1 = i + i0 ∧ 0 < i0 ⊦ i + i2 < i1 + i3

Rule argument

spec: int-basic2 name: ex-ls-add seq: ⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0) subst: [j, i] → [i1, i] flags: <all paths>

Additional information

Used terms: i1, i