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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# bits2 ≤ # bits1, ∀ bits1, bits0. # bits0 < #(bits2 + true ') → # bits0 ≤ 1 + # bits1 → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bits1 = [], bits2 = [], bitsor(bits2, true ' + bits1.butlast) = bits2, false ∈ bits2

Rule argument

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;