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Proof step 3 from proof fits-law-bxor-02 (previous step)

⊦ bor(band(s2i(sho1), bcompl(s2i(sho2))), band(bcompl(s2i(sho1)), s2i(sho2))) = s2i(i2s(bor(band(s2i(sho1), bcompl(s2i(sho2))), band(bcompl(s2i(sho1)), s2i(sho2)))))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ∈short ∧ j ∈short → s2i(i2s(bor(i, j))) = bor(i, j)
Proof step 5:
¬(band(s2i(sho1), bcompl(s2i(sho2))) ∈short ∧ band(bcompl(s2i(sho1)), s2i(sho2)) ∈short) ⊦ bor(band(s2i(sho1), bcompl(s2i(sho2))), band(bcompl(s2i(sho1)), s2i(sho2))) = s2i(i2s(bor(band(s2i(sho1), bcompl(s2i(sho2))), band(bcompl(s2i(sho1)), s2i(sho2)))))

Rule argument

name: bor-sis rotate: F subst: [j, i] → [band(bcompl(s2i(sho1)), s2i(sho2)), band(s2i(sho1), bcompl(s2i(sho2)))]

Additional information

Used terms: band(bcompl(s2i(sho1)), s2i(sho2)), band(s2i(sho1), bcompl(s2i(sho2)))