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Proof step 110 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 111:
# bits0 = 1, # bits = i→n(j), int2bits(i) = bits0 + bits, bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))) = bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))), 0 ≤ j, 0 ≤ i, 2 ^ j * 2 < i ⊦ bits = [], bits0 = [], bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))) = [], (sgn(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))))[0] ↔ ¬ bits0.last) ∧ sgn(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))))[0, bits0.last] = bits0 + bits

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);