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Proof step 43 from proof bxor-flip-one-bit (previous step)

# bits0 - 1 < # bits0, int2bits(i) = bits0 + bits, # bits = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits0 = [], (int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1] ↔ ¬ bits0[# bits0 - 1]) ∧ int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1, bits0[# bits0 - 1]] = bits0 + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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