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Proof step 57 from proof bxor-flip-one-bit (previous step)

int2bits(i) = bits + boolvar ', 2 < i, 0 < i, boolvar ∈ int2bits(i) ⊦ i = 0, (sgn(bitsxor(bits, false ') + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(bitsxor(bits, false ') + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 58:
⊦ bits ≠ [] ∧ bits0 = false ' → bitsxor(bits, bits0) = bits
Proof step 59:
¬(bits ≠ [] ∧ false ' = false '), int2bits(i) = bits + boolvar ', 2 < i, 0 < i, boolvar ∈ int2bits(i) ⊦ i = 0, (sgn(bitsxor(bits, false ') + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(bitsxor(bits, false ') + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)
Proof step 62:
bits ≠ [] ∧ false ' = false ', int2bits(i) = bits + boolvar ', 2 < i, 0 < i, boolvar ∈ int2bits(i) ⊦ i = 0, (sgn(bits + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)

Rule argument

name: bitsxor-rec-02 rotate: F subst: [bits0, bits] → [false ', bits]

Additional information

Used terms: false ', bits