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Proof step 19 from proof bitsor-sgn-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
true ∈ bits ⊦ bits = [], boolvar ↔ boolvar ∈ bitsor(bits, true ' + bits1), false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;