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Proof step 15 from proof bitsand-zero-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
bits.last, bits0.last, false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], # bits = 1, # bits0 = 1, false ∈ bitsand(bits.butlast, bits0.butlast)

Rule argument

Additional information

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