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Proof step 95 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ # bits1 ≤ # bits0 ∧ bits1 ≠ [] ∧ ¬ bits1.first → bitsand(bits + bits0, bits1) = bitsand(bits0, bits1) subst: [bits1, bits0, bits] → [fillsgn(8, int2bits(i)), bits, bits0] rotate: F spec: intops inst: name: bitsand-app-false

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ 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, bits0