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Proof step 1 from proof short-law-mult (conclusion)

⊦ 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 2:
⊦ s2i(i2s(j * j0)) = s2i(i2s(s2i(i2s(j)) * j0))
Proof step 3:
⊦ s2i(i2s(s2i(i2s(j)) * j0)) = s2i(i2s(s2i(i2s(j)) * s2i(i2s(j0))))

Rule argument

name: short-law-mult-one rotate: F subst: [j0, j] → [j0, j]

Additional information

Used terms: j0, j