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Proof step 16 from proof bitsor-ones-false (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ bits1, bits0. # bits0 < 1 → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bitsor(true ', false ' + bits1) = true '

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ # boolvar ' = 0 +1;% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)