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Proof step 1 from proof le-add-01 (conclusion)

⊦ i + j ≤ i0 ∧ j ≥ 0 ∧ j0 ≥ 0 → i - j0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j0 + i0 < i ⊦ i0 < i + j, j < 0, j0 < 0

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%ge-def: ⊦ i ≥ j ↔ ¬ i < j;%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0;