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Proof step 3 from proof short-law-add (previous step)

⊦ s2i(i2s(s2i(i2s(j)) + j0)) = s2i(i2s(s2i(i2s(j)) + s2i(i2s(j0))))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i + j = j + i
Proof step 5:
⊦ s2i(i2s(j0 + s2i(i2s(j)))) = s2i(i2s(s2i(i2s(j)) + s2i(i2s(j0))))

Rule argument

seq: ⊦ i + j = j + i subst: [j, i] → [j0, s2i(i2s(j))] rotate: F spec: int-basic2 inst: name: add-c

Additional information

Used terms: j0, s2i(i2s(j))