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Proof step 2 from proof bor-bib (previous step)

~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ b2i(i2b(bor(i, j))) = bor(i, j)

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ bor(b2i(by1), b2i(by2)) = b2i(i2b(bor(b2i(by1), b2i(by2))))

Rule argument

name: fits-law-bor subst: [by1, by2] → [i2b(i), i2b(j)]

Additional information

Used simplifier rules: bib-in-01: ⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i; Used terms: i2b(i), i2b(j)