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Proof step 84 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, int2bits(i)), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), 0 ≤ j, # int2bits(i) + 1 < # bitsand(fillsgn(8, int2bits(i)), int2bits(j)), # int2bits(j) < # bitsand(fillsgn(8, int2bits(i)), int2bits(j)), # int2bits(j) + 1 < # bitsand(fillsgn(8, int2bits(i)), int2bits(j)), 8 < # bitsand(fillsgn(8, int2bits(i)), int2bits(j)), # int2bits(i) < 8, # int2bits(j) < 8, # int2bits(i) < # bitsand(fillsgn(8, int2bits(i)), int2bits(j)) ⊦ bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = [], 8 = # fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j))), fillsgn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 85:
⊦ ¬ 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(j), fillsgn(8, int2bits(i))] rotate: F spec: intops inst: name: bitsand-len-false

Additional information

Used simplifier rules: ⊦ ¬ n < n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) Used terms: int2bits(j), fillsgn(8, int2bits(i))