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Proof step 15 from proof bitsand-onein (previous step)

false ' + bits0 ≠ [] ∧ bits1 = false ', ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ false ', bits1 = [], bits1 = true ', true ∈ bits, bits0 ≠ [] ∧ # bits1 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
true ∈ bitsand(bits + false ' + bits0, false '), ∀ bits0, bits, bits2. # bits2 < # false ' → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bits

Rule argument

Additional information

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