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Proof step 12 from proof fits-law-bxor-02 (previous step)

¬(s2i(sho1) ∈short ∧ bcompl(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))))), s2i(i2s(band(s2i(sho1), bcompl(s2i(sho2))))) = band(s2i(sho1), bcompl(s2i(sho2))) ∧ s2i(i2s(band(bcompl(s2i(sho1)), s2i(sho2)))) = band(bcompl(s2i(sho1)), s2i(sho2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ 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))))), bcompl(s2i(sho2)) ∈short, s2i(i2s(band(s2i(sho1), bcompl(s2i(sho2))))) = band(s2i(sho1), bcompl(s2i(sho2))) ∧ s2i(i2s(band(bcompl(s2i(sho1)), s2i(sho2)))) = band(bcompl(s2i(sho1)), s2i(sho2))

Rule argument

Additional information

Used simplifier rules: short2int-in: ⊦ s2i(sho) ∈short;