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Proof step 6 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, bits0) ↔ boolvar ∈ bits0, bits0 = [], bits = [], false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0, bits0 = [], bits = [], false ∈ bits, bits ≠ bits.first ' ∧ bits0 ≠ bits0.first '

Rule argument

Additional information

Used simplifier rules: ⊦ (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);