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Proof step 9 from proof le-add-08 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 10:
0 < i, i1 + i2 + i ≤ j, i0 + i1 + i2 + i ≤ j, 0 ≤ i0, i1 + i2 ≤ j ⊦ i1 + i2 ≤ i0 + 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% ⊦ i < j → ¬ j < i; (s)