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Proof step 154 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 155:
int2bits(i) = false ' + (false ' + bits1) + boolvar ' + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = false ' + (false ' + bits1) + (¬ boolvar)' + bitsxor(bits, mklist(false, i→n(j))), 0 < i, 0 ≤ j, 3 + i→n(j) ≤ # int2bits(i), false ∈ int2bits(i), false ∈ bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) ⊦ i = 0, # int2bits(i) = 2 + i→n(j), i = 2 ^ j * 2, bits = [], (sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[2 + # bits1] ↔ ¬ boolvar) ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[2 + # bits1, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;