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Proof step 23 from proof bitsor-first-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
bits ≠ [] ⊦ bits0 + boolvar ' = bits ↔ bits0 = bits.butlast ∧ (boolvar ↔ bits.last)
Proof step 25:
true ' + bits1 ≠ [], bitsor(false ' + bits3, false ' + bits0) = (true ' + bits1).butlast ∧ (boolvar ∨ boolvar0 ↔ (true ' + bits1).last), ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0

Rule argument

seq: bits ≠ [] ⊦ bits0 + boolvar ' = bits ↔ bits0 = bits.butlast ∧ (boolvar ↔ bits.last) subst: [bits, boolvar, bits0] → [true ' + bits1, boolvar ∨ boolvar0, bitsor(false ' + bits3, false ' + bits0)] rotate: F spec: list-last inst: bitlista name: splitlasteq

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: true ' + bits1, boolvar ∨ boolvar0, bitsor(false ' + bits3, false ' + bits0)