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Proof step 24 from proof and-zerosin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
∀ bits1, bits0. # bits1 < # bits → true ∈ bitsand(bits1, bits0) → bits1 = [] ∨ bits0 = [] ∨ true ∈ bits1 ⊦ bits = [], bits = bits.first ', bits2 + boolvar ' = (bits2 + boolvar ').first ', true ∈ bits, ¬ true ∈ bitsand(bits.butlast, bits2) ∧ (¬ bits.last ∨ ¬ boolvar)

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];