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

bits1 = bits + boolvar ', # bits2 ≤ # bits1, ∀ bits1, bits0. # bits0 < #(bits2 + true ') → # bits0 ≤ 1 + # bits1 → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bits1 = [], bits2 = [], bitsor(bits2, true ' + bits) = bits2, false ∈ bits2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# bits2 ≤ # bits + 1, ∀ bits1, bits0. # bits0 < #(bits2 + true ') → # bits0 ≤ 1 + # bits1 → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bits2 = [], bitsor(bits2, true ' + bits) = bits2, false ∈ bits2

Rule argument

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];