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Proof step 3 from proof bor-assoc (previous step)

⊦ bor(bcompl(band(bcompl(i), bcompl(i0))), j0) = bor(i, bor(i0, j0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ bor(i, j) = bcompl(band(bcompl(i), bcompl(j)))
Proof step 5:
⊦ bcompl(band(bcompl(bcompl(band(bcompl(i), bcompl(i0)))), bcompl(j0))) = bor(i, bor(i0, j0))

Rule argument

name: bor-notAndNot rotate: F subst: [j, i] → [j0, bcompl(band(bcompl(i), bcompl(i0)))]

Additional information

Used terms: j0, bcompl(band(bcompl(i), bcompl(i0)))