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Proof step 5 from proof abssub-05 (previous step)

i = i1 + j, j < i0 ⊦ abs(i1) + j = i, i + 1 < i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j < i0, i1 < 0 ⊦ (i1 + j) + 1 < i0

Rule argument

Additional information

Used simplifier rules: abs-18: ⊦ abs(i) = i ↔ 0 ≤ i;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2