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Proof step 10 from proof le-add-08 (previous step)

0 < i, i1 + i2 + i ≤ j, i0 + i1 + i2 + i ≤ j, 0 ≤ i0, i1 + i2 ≤ j ⊦ i1 + i2 ≤ i0 + j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i + j = j + i
Proof step 12:
0 < i, i1 + i2 ≤ j, 0 ≤ i0, i0 + i1 + i2 + i ≤ j, i1 + i2 + i ≤ j ⊦ i1 + i2 ≤ j + i0

Rule argument

spec: int-basic2 name: add-c seq: ⊦ i + j = j + i subst: [j, i] → [j, i0] flags: <all paths>

Additional information

Used terms: j, i0