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Proof step 5 from proof short-law-add (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s2i(i2s(j + j0)) = s2i(i2s(s2i(i2s(j)) + j0))

Rule argument

name: short-law-add-one rotate: F subst: [j0, j] → [s2i(i2s(j)), j0]

Additional information

Used simplifier rules: ⊦ i + j = j + i; Used terms: s2i(i2s(j)), j0