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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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