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Proof step 116 from proof byte-law-and-one (previous step)

7 = # fillsgn(7, int2bits(i)), int2bits(j) = bits0 + bits, # bits = 8, # int2bits(i) < # bitsand(bits, int2bits(i)), 0 ≤ i, # int2bits(i) < 8, 8 < # bitsand(bits, int2bits(i)), # int2bits(i) + 1 < # bitsand(bits, int2bits(i)) ⊦ lastn(8, bitsand(bits, int2bits(i))) = lastn(8, fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i))))), bits0 + bits = [], 8 = # fillsgn(8, bitsand(bits, int2bits(i))), bitsand(bits, int2bits(i)) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ ¬ n < n;% ⊦ 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(i), bits