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Proof step 21 from proof bitsor-first-02 (previous step)

bitsor(false ' + bits3, false ' + bits2.butlast) + (boolvar ∨ bits2.last)' = true ' + bits1, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits2 = [], ¬ boolvar ∧ ¬ bits2.last ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits2.butlast) ∧ ¬ boolvar ∧ ¬ bits2.last

User applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
bits2 = bits0 + boolvar0 ', bitsor(false ' + bits3, false ' + bits0) + (boolvar ∨ boolvar0)' = true ' + bits1, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits2 = [], ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0

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