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Proof step 24 from proof bytes2int-len-one-01 (previous step)

b2i(by) < 128, 0 ≤ b2i(by) ⊦ by0 = i2b(0), true ∈ bytes2bits(bytes), true ∈ fillsgn(8, int2bits(b2i(by0)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits

Rule argument

seq: ⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits subst: [bits, n] → [int2bits(b2i(by0)), 8] rotate: F spec: bitopsaux inst: name: fill-fill

Additional information

Used simplifier rules: ⊦ ¬ 8 < # int2bits(b2i(by));% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ int2bits(j).first ↔ j < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ true ∈ int2bits(j) ↔ j ≠ 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ b2i(by) = i → (i2b(i) ≠ by ↔ false); Used terms: int2bits(b2i(by0)), 8