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Proof step 9 from proof rax-03 (previous step)

~ i0 = i1 + 1, i0 < 0 ⊦ i→n(i1) = i→n(i0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i0 + i1 + 1 = 0, i0 < i1, i0 + i1 < 0, i0 < 0, 0 ≤ i1 ⊦ i→n(i1) = i→n(i0 + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j < j ↔ i < 0; (s)% ⊦ i ≤ i; (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i ≤ i + j ↔ 0 ≤ j; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-03: ⊦ i + 1 = 0 → i < 0; (forward) from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)