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Proof step 18 from proof bitsor-zeroes-true (previous step)

bits1 = bits + boolvar ', ∀ bits1, bits0. # bits0 < #(bits2 + false ') → true → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = true ' + bits1 ∨ true ∈ bits0 ⊦ bits1 = [], bits2 = [], bitsor(bits2, true ' + bits) + boolvar ' = true ' + bits1, true ∈ bits2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ bits1, bits0. # bits0 < #(bits2 + false ') → true → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = true ' + bits1 ∨ true ∈ bits0 ⊦ bits2 = [], bitsor(bits2, true ' + bits) = true ' + bits, true ∈ bits2

Rule argument

Additional information

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