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Proof step 22 from proof and-zerosin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
∀ 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, bits0.butlast) ∧ (¬ bits.last ∨ ¬ bits0.last)

Rule argument

Additional information

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