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Proof step 1 from proof abs-18 (conclusion)

⊦ abs(i) = i ↔ 0 ≤ i

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero: ⊦ i < 0 ∧ j < 0 → i + j ≠ 0; from specification int-basic2