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Proof step 151 from proof byte-law-and-one (previous step)

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

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] → [fillsgn(8, int2bits(i)), bits] rotate: F spec: intops inst: name: bitsand-len-true

Additional information

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