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Proof step 1 from proof absp (conclusion)

0 ≤ j, j < i ⊦ abs(j - i) ≤ abs(i)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-16: ⊦ 0 < j → abs(j) = j;%abs-ls-03: ⊦ j < i → (i < abs(j - i) ↔ j < 0);%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lf-04: ⊦ ¬ j < i ∧ j < i0 → i < i0; (forward) from specification int-basic1