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Proof step 1 from proof abs-sub-02 (conclusion)

⊦ i < j → ¬ abs(j - i) ≤ abs(j - i + 1)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abssub-02: ⊦ j < i → abs(i - j) = i - j;%abssub-01: ⊦ j < i → abs(i - j + 1) = i - j + 1;%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-add-05: ⊦ i + j - i + 1 = j - 1;%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;%succ-01: ⊦ i + 1 ≤ j ↔ i < j;%irref: ⊦ ¬ i < i; 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