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Proof step 70 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < 8, 8 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) ≤ # int2bits(i), # int2bits(j) < 8 ⊦ lastn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits subst: [bits0, bits] → [int2bits(j), int2bits(i)] rotate: F spec: intops inst: name: bitsand-len-true

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ int2bits(j).first ↔ j < 0;% ⊦ ¬ n < n; Used terms: int2bits(j), int2bits(i)