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Proof step 44 from proof bitsor-sgn-in (previous step)

bits0 = boolvar0 + bits1 ⊦ bits = [], boolvar ∈ bits0 ↔ boolvar ∈ bitsor(bits, bits0), bits0 = [], true ∈ bits, boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
⊦ bits = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits, false ' + bits1), true ∈ bits

Rule argument

Additional information

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