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Proof step 1 from proof bor-assoc (conclusion)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bor(i, j) = bcompl(band(bcompl(i), bcompl(j)))
Proof step 3:
⊦ bor(bcompl(band(bcompl(i), bcompl(i0))), j0) = bor(i, bor(i0, j0))

Rule argument

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

Additional information

Used terms: i0, i