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Proof step 3 from proof le-add-04 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i ≤ i0 ⊦ i + j0 ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0

Rule argument

pos: left 1

Additional information

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