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Proof step 4 from proof le-add-08 (previous step)

0 < i, i0 + j < i1 + i2, i0 + i1 + i2 + i ≤ j, 0 ≤ i0, (i1 + i2 + i) + i0 ≤ j ∧ i0 ≥ 0 → i1 + i2 + i ≤ j ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 5:
i0 + j < i1 + i2, 0 < i, 0 ≤ i0, i0 + i1 + i2 + i ≤ j, i1 + i2 + i ≤ j ⊦

Rule argument

pos: left 5

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