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Proof step 11 from proof bitsor-zeroes-true (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# bits = 1, ∀ bits1, bits0. # bits0 < 1 → true → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = true ' + bits1 ∨ true ∈ bits0 ⊦ bitsor(bits, true ' + bits1) = true ' + bits1, bits = [], bits1 = [], true ∈ bits

Rule argument

Additional information

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