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Proof step 1 from proof bitsxor-bit (conclusion)

⊦ bitsxor(boolvar ', boolvar0 ') = (¬(boolvar ↔ boolvar0))'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsxor(bits, bits0) = bitsinvert(bits)
Proof step 3:
¬(boolvar ' ≠ [] ∧ boolvar0 ' = true ') ⊦ bitsxor(boolvar ', boolvar0 ') = (¬(boolvar ↔ boolvar0))'

Rule argument

name: bitsxor-rec-01-01 rotate: F subst: [bits0, bits] → [boolvar0 ', boolvar ']

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ bitsinvert(boolvar ') = (¬ boolvar)'; Used terms: boolvar0 ', boolvar '