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Proof step 18 from proof and-false (previous step)

false ' + bits1 ≠ [], 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 ⊦ 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: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];