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Proof step 36 from proof byte-law-and-one (previous step)

bitsand(lastn(8, fillsgn(8, int2bits(i))), int2bits(j)) = [] ⊦ lastn(8, fillsgn(8, sgn(bitsand(int2bits(i), int2bits(j))))) = ((((((false ' + false ') + false ') + false ') + false ') + false ') + false ') + false '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 37:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ []

Rule argument

seq: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [] subst: [bits0, bits] → [int2bits(j), lastn(8, fillsgn(8, int2bits(i)))] rotate: F spec: intops inst: name: and-notempty

Additional information

Used simplifier rules: ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ int2bits(j) ≠ []; Used terms: int2bits(j), lastn(8, fillsgn(8, int2bits(i)))