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Proof step 1 from proof absls-05 (conclusion)

⊦ abs(i) ≤ abs(i -1) ↔ i ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-09: ⊦ 0 < i → abs(i - 1) = abs(i) - 1;%abs-16: ⊦ 0 < j → abs(j) = j;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%succ-02: ⊦ i < j + 1 ↔ i ≤ j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1