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Proof step 4 from proof bitsxor-rec-01 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
bits0 = bits + boolvar0 ' ⊦ # bits0 = 1, bits0 = [], bitsxor(bits1 + boolvar ', bits0) = bitsxor(bits1, bits) + (¬(boolvar ↔ boolvar0))', bits1 = []

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits0, boolvar0, bits] 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: bits0, boolvar0, bits