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Proof step 20 from proof bitsor-first-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];