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Proof step 10 from proof ple-01 (previous step)

∃ i1. j0 = i0 + i1 ∧ 0 < i1, i1 + 1 < j ⊦ (i1 + 1) + i0 < j + j0, i1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2