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Proof step 2 from proof pls-01 (previous step)

j < j0, i ≤ j + i0 ⊦ i < j0 + i0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%lf-04: ⊦ ¬ j < i ∧ j < i0 → i < i0; (forward) from specification int-basic1%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) 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% ⊦ i + j < i + i0 ↔ j < i0; (s)%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1