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Proof step 25 from proof and-onesin (previous step)

bits0 = bits2 + boolvar0 ', ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits0 = [], bits = [], (boolvar ↔ bits.last ∧ boolvar0) ∨ boolvar ∈ bitsand(bits.butlast, bits2) ↔ boolvar ∈ bits0, bits = bits.first ', bits0 = bits0.first ', false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits = bits.first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits = [], (boolvar ↔ bits.last ∧ boolvar0) ∨ boolvar ∈ bitsand(bits.butlast, bits2) ↔ (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];