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Proof step 5 from proof le-add-06 (previous step)

(i0 + i1) + i ≤ i2 ∧ i ≥ 0 → i0 + i1 ≤ i2 ⊦ 0 ≤ i ∧ i0 + i + i1 ≤ i2 → ¬ ¬ i1 + i0 ≤ i + i2

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%ge-def: ⊦ i ≥ j ↔ ¬ i < j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2