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Proof step 12 from proof bitsxor-append-same-size (previous step)

# bits2 = 1, ∀ bits1, bits, bits2, bits3. # bits3 < 1 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bits = [], bits1 = [], bits2 = [], bitsxor(bits, bits1 + bits2.butlast) + (¬(boolvar ↔ bits2.last))' = bitsxor(bits, bits1) + bitsxor(boolvar ', bits2)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
bits2 = bits0 + boolvar0 ', # bits2 = 1, ∀ bits1, bits, bits2, bits3. # bits3 < 1 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bits = [], bits1 = [], bits2 = [], bitsxor(bits, bits1 + bits0) + (¬(boolvar ↔ boolvar0))' = bitsxor(bits, bits1) + bitsxor(boolvar ', bits2)

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits2, boolvar0, bits0] 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: bits2, boolvar0, bits0