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Proof step 3 from proof bytes2bits-mklist-01 (previous step)

bytes2bits(mklist(i2b(~1), m)) = mklist(true, 8 * m) ⊦ byte2bits(i2b(~1)) = true ' + true ' + true ' + true ' + true ' + true ' + true ' + true '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ byte2bits(by) = fillsgn(8, int2bits(b2i(by)))

Rule argument

name: byte2bits rotate: F subst: [by] → [i2b(~1)]

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;%i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);% ⊦ uint2bits(1) = false + true;% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ sgn(true ' + true ') = true ';% ⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n);%n ≠ 0 ⊦ mklist(true, (* n)) = true ' + mklist(true, n -1);% ⊦ mklist(boolvar, 0) = [];% ⊦ bits + [] = bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1; Used terms: i2b(~1)