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Proof step 67 from proof byte-law-and-one (previous step)

6 = # fillsgn(6, int2bits(i)), # int2bits(j) < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < 8, 8 < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < 8, # int2bits(i) < # int2bits(j), # int2bits(i) < 7, # int2bits(i) + 1 < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) + 1 < # bitsand(int2bits(j), int2bits(i)), 0 ≤ i ⊦ 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i))), lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand(false ' + false ' + fillsgn(6, int2bits(i)), int2bits(j))))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 68:
⊦ ¬ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits0

Rule argument

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

Additional information

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