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Proof step 1 from proof absls-03 (conclusion)

⊦ i + 1 < abs(i) ↔ i < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (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%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2