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Proof step 8 from proof div-shift-two (previous step)

j0 < j, 0 < j, 0 ≤ i0, 0 ≤ j0 ⊦ i0 = 2 * i0 / 2 + (i0 * j + j0 < j * (2 * i0 / 2 + 1) ⊃ 0;1), j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i + j = j + i
Proof step 10:
0 < j, j0 < j, 0 ≤ j0, 0 ≤ i0 ⊦ i0 = 2 * i0 / 2 + (i0 * j + j0 < j * (1 + 2 * i0 / 2) ⊃ 0;1), j = 0

Rule argument

spec: int-basic2 name: add-c seq: ⊦ i + j = j + i subst: [j, i] → [1, 2 * i0 / 2] flags: <all paths>

Additional information

Used terms: 1, 2 * i0 / 2