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Proof step 1 from proof bor-zero (conclusion)

⊦ bor(i, 0) = i

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:
⊦ bcompl(band(bcompl(i), bcompl(0))) = i

Rule argument

name: bor-notAndNot rotate: F subst: [j, i] → [0, i]

Additional information

Used terms: 0, i