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Proof step 131 from proof bxor-flip-one-bit (previous step)

bits0 = bits1 + boolvar ', int2bits(i) = bits0 + bits, # bits = i→n(j), bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))) = (bitsxor(bits1, false ') + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits0 = [], bits = [], # bits0 = 1, (sgn(bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))))[# bits0 - 1] ↔ ¬ boolvar) ∧ sgn(bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))))[# bits0 - 1, boolvar] = bits0 + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 132:
int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bitsxor(bits1, false ') + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], (sgn((bitsxor(bits1, false ') + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bitsxor(bits1, false ') + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];