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Proof step 64 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)) ⊦ lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand((¬ 0 ≤ i)' + (¬ 0 ≤ i)' + fillsgn(6, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 65:
⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits
Proof step 66:
¬(int2bits(i).first ∧ # int2bits(i) ≤ # int2bits(j) ∧ int2bits(i) ≠ []), 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)) ⊦ lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand((¬ 0 ≤ i)' + (¬ 0 ≤ i)' + fillsgn(6, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i)))

Rule argument

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

Additional information

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