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Proof step 2 from proof bitsxor-empty (previous step)

bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsxor(bits, bits0) = bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last))
Proof step 4:
¬(1 < # bits ∧ 1 < # bits0), bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = []

Rule argument

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

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits0, bits