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Proof step 2 from proof int-add-pos-01 (previous step)

j + i = j1, j < j1, i ≤ 0, 0 ≤ j ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < i + j ↔ 0 < j; (s)