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Proof step 20 from proof and-zerosin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
bits0 = bits0.first ', true ∈ bitsand(false ' + bits2, bits0), ∀ bits1, bits0. # bits1 < #(false + bits2) → true ∈ bitsand(bits1, bits0) → bits1 = [] ∨ bits0 = [] ∨ true ∈ bits1 ⊦ false ' + bits2 = true ', bits0 = false ', bits0 = true ', bits0 = [], bits2 = [], true ∈ bits2

Rule argument

Additional information

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