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Proof step 5 from proof bor-assoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ bcompl(band(bcompl(j0), band(bcompl(i), bcompl(i0)))) = bor(i, bor(i0, j0))

Rule argument

Additional information

Used simplifier rules: bcompl-absorption: ⊦ bcompl(bcompl(i)) = i;