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Proof step 20 from proof and-false (previous step)

bitsand(true ' + bits3, true ' + bits2) = (false ' + bits1).butlast, boolvar ∧ boolvar0 ↔ (false ' + bits1).last, ∀ bits1, bits, bits0. # bits0 < #(bits3 + boolvar ') → bitsand(true ' + bits0, true ' + bits) = false ' + bits1 → false ⊦ bits1 = [], 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 ' + bits).last ↔ bits.last);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;