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Proof step 1 from proof bitsxor-rec-01 (conclusion)

⊦ 1 < # bits ∧ 1 < # bits0 → bitsxor(bits, bits0) = bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bitsxor(bits, bits0) = bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last))', # bits = 1, bits = [], # bits0 = 1, bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ bits + boolvar = bits + boolvar ';