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Proof step 1 from proof absls-06 (conclusion)

⊦ i < 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;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-01: ⊦ i < 0 ∧ j < 0 → i + j < 0; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1