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Proof step 144 from proof bxor-flip-one-bit (previous step)

bits1 = boolvar0 + bits0, int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 0 ≤ i, 0 ≤ j, 3 + i→n(j) ≤ # int2bits(i), 0 < i ⊦ bits = [], bits1 = [], boolvar0 ↔ ¬ (bits0 + (¬ boolvar)').first, i = 2 ^ j * 2, (sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 145:
bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + bits0) + (¬ boolvar)' + bitsxor(bits, mklist(false, i→n(j))), # bits = i→n(j), int2bits(i) = (false ' + bits0) + boolvar ' + bits, 0 < i, 3 + i→n(j) ≤ # int2bits(i), 0 ≤ j, 0 ≤ i, false ∈ bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))), false ∈ int2bits(i) ⊦ bits = [], i = 2 ^ j * 2, (bits0 + (¬ boolvar)').first, (sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0] ↔ ¬ boolvar) ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(i) = boolvar ' + bits ∧ 0 ≤ i → ¬ boolvar; (forward)% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ m + n = n + m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ 0 < j → int2bits(j) ≠ true ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;