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Proof step 2 from proof bitsxor-rec-01 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
bits = bits1 + boolvar ' ⊦ bitsxor(bits, bits0) = bitsxor(bits1, bits0.butlast) + (¬(boolvar ↔ bits0.last))', # bits = 1, bits = [], # bits0 = 1, bits0 = []

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits, boolvar, bits1] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits, boolvar, bits1