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Proof step 139 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits) subst: [bits, n] → [int2bits(i), 8] rotate: F spec: bitopsaux inst: name: fill-rec

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ int2bits(j) ≠ [];% ⊦ int2bits(j).first ↔ j < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: int2bits(i), 8