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Proof step 1 from proof abs-ls-04 (conclusion)

0 ≤ i, abs(j) < i ⊦ 0 ≤ i + j

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%Abs-neg: ⊦ j < 0 → abs(j) = 0 - j;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%ls-01: ⊦ i < j → ¬ j < 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%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1