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Proof step 23 from proof and-onesin (previous step)

1 < # bits ∧ 1 < # bits0, ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last) ↔ boolvar ∈ bits0, bits0 = [], bits = [], false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits0 = [], bits = [], (boolvar ↔ bits.last ∧ bits0.last) ∨ boolvar ∈ bitsand(bits.butlast, bits0.butlast) ↔ boolvar ∈ bits0, bits = bits.first ', bits0 = bits0.first ', false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar = bits + boolvar ';