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Proof step 50 from proof bitsor-bi-sgn (previous step)

¬ false ∈ true ' + bits0 ∧ true ' + bits0 ≠ [] ∧ bits3 ≠ [], ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ ~1 = bits2int(true ' + true '), bits3 = [], false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ ~1 = bits2int(true ' + true '), bits3 = [], false ∈ bits0

Rule argument

Additional information

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