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Proof step 21 from proof bitsor-zeroes-false (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# bits2 ≤ # bits + 1, ∀ bits1, bits0. # bits0 < #(bits2 + false ') → # bits0 ≤ 1 + # bits1 → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = false ' + bits1 ∨ true ∈ bits0 ⊦ bits2 = [], bitsor(bits2, false ' + bits) = false ' + bits, true ∈ bits2

Rule argument

Additional information

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