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Proof step 160 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 161:
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))), bits1.first ↔ ¬ (bits1.rest + (¬ boolvar)').first, 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], (bits1 + (¬ boolvar)')[# bits1, boolvar] + bitsxor(bits, mklist(false, i→n(j))) = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;