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Proof step 1 from proof sgn-04 (conclusion)

⊦ j ≤ i → sgn(i - j) = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sgn-01: ⊦ sgn(i) = 1 ↔ 0 ≤ i;%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs