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Proof step 55 from proof bitsor-sgn-in (previous step)

# bits0 = # bits1 ⊦ bits0 ≠ [], bits2 + bits0 = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits2 + bits0, false ' + bits1), # bits2 = 1, bits2 = [], true ∈ bits2, true ∈ bits0, bits0 ≠ [] ∧ bits1 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
⊦ bits2 = [], # bits2 = 1, ¬ boolvar ↔ boolvar ∈ bitsor(bits2, false '), true ∈ bits2

Rule argument

Additional information

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