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Proof step 15 from proof fits-law-bor-01 (previous step)

s2i(sho1) ∈short ⊦ s2i(i2s(band(bcompl(s2i(sho1)), bcompl(b2i(by2))))) = band(bcompl(s2i(sho1)), bcompl(b2i(by2))), bcompl(band(bcompl(s2i(sho1)), bcompl(b2i(by2)))) = s2i(i2s(bcompl(band(bcompl(s2i(sho1)), bcompl(b2i(by2)))))), bcompl(s2i(sho1)) = bcompl(s2i(sho1)) ∧ bcompl(b2i(by2)) ∈short

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ s2i(i2s(band(bcompl(s2i(sho1)), bcompl(b2i(by2))))) = band(bcompl(s2i(sho1)), bcompl(b2i(by2))), bcompl(band(bcompl(s2i(sho1)), bcompl(b2i(by2)))) = s2i(i2s(bcompl(band(bcompl(s2i(sho1)), bcompl(b2i(by2)))))), bcompl(b2i(by2)) ∈short

Rule argument

Additional information

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