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Proof step 30 from proof bitsor-first-02 (previous step)

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 ⊦ bits1 = [], ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
bitsor(false ' + bits3, false ' + bits0) = true ' + bits1.butlast, boolvar ∨ boolvar0 ↔ bits1.last, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits1 = [], ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits ≠ [] → ((boolvar ' + bits).last ↔ bits.last);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;