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Proof step 3 from proof cast-law-or-one (previous step)

0 < i ⊦ cast(i, bcompl(band(bcompl(j), bcompl(j0)))) = cast(i, bor(cast(i, j), 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:
0 < i ⊦ cast(i, bcompl(band(bcompl(j), bcompl(j0)))) = cast(i, bcompl(band(bcompl(cast(i, j)), bcompl(j0))))

Rule argument

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

Additional information

Used terms: j0, cast(i, j)