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Proof step 17 from proof and-notone-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits.last ∈ bits ↔ bits ≠ [];% ⊦ 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;