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Proof step 17 from proof ple-01 (previous step)

i1 + 1 < j, 0 < i ⊦ i1 < j, i1 + 1 < j + i

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-basic2 name: ex-ls-add seq: ⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0) subst: [j, i] → [j, i1 + 1] flags: <all paths>

Additional information

Used simplifier rules: succ-02: ⊦ i < j + 1 ↔ i ≤ j;%le-def: ⊦ i ≤ j ↔ ¬ j < i;% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, i1 + 1