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Proof step 6 from proof bor-assoc (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ bor(i, j) = bcompl(band(bcompl(i), bcompl(j)))
Proof step 8:
⊦ bcompl(band(bcompl(j0), band(bcompl(i), bcompl(i0)))) = bor(i, bcompl(band(bcompl(i0), bcompl(j0))))

Rule argument

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

Additional information

Used terms: j0, i0