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Proof step 15 from proof and-false (previous step)

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

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 ' ≠ [];