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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

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