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Proof step 18 from proof bxor-flip-one-bit (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 19:
# bits0 = 0, int2bits(i) = bits0 + bits, # bits = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ (int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1] ↔ ¬ (bits0 + bits)[# bits0 - 1]) ∧ int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1, (bits0 + bits)[# bits0 - 1]] = bits0 + bits
Proof step 40:
int2bits(i) = bits0 + bits, # bits = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ # bits0 = 0, (int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1] ↔ ¬ (bits0 + bits)[# bits0 - 1]) ∧ int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1, (bits0 + bits)[# bits0 - 1]] = bits0 + bits

Rule argument

fma: # bits0 = 0

Additional information

Used formulas: # bits0 = 0