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Proof step 12 from proof and-notone-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
bitsand(boolvar ' + bits3.butlast, false ' + bits0.butlast) + (bits0.last ∧ bits3.last)' = true ' + bits1, ∀ bits2, bits0, bits1. # bits2 < #(boolvar + bits3) → bitsand(bits2, false ' + bits0) = true ' + bits1 → bits2 = [] ⊦ 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)

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)% ⊦ boolvar ' + bits ≠ [];% ⊦ bits ≠ [] → ((boolvar ' + bits).last ↔ bits.last);% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ boolvar + bits = boolvar ' + bits;