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Proof step 2 from proof int-simp-04 (previous step)

i < i1, i2 < i3, i1 + i3 < i + i2 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: i1, i