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Proof step 5 from proof bxor-flip-one-bit (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 6:
n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n
Proof step 7:
2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ i→n(j) ≤ # int2bits(i), ( int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j)))))[# int2bits(i) - i→n(j) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(j) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))) [# int2bits(i) - i→n(j) + 1, int2bits(i)[# int2bits(i) - i→n(j) + 1] ] = int2bits(i)
Proof step 17:
∃ bits0, bits1. int2bits(i) = bits0 + bits1 ∧ # bits1 = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ ( int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j)))))[# int2bits(i) - i→n(j) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(j) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))) [# int2bits(i) - i→n(j) + 1, int2bits(i)[# int2bits(i) - i→n(j) + 1] ] = int2bits(i)

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [i→n(j), int2bits(i)] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used terms: i→n(j), int2bits(i)