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Proof step 20 from proof bitsxor-zeros (previous step)

bits0 = bits3 + boolvar0 ', # bits0 ≤ # bits2 + 1, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ # bits0 = 1, bits0 = [], bitsxor(bits2, bits3) + (¬(boolvar ↔ boolvar0))' = bits2 + boolvar ', bits2 = [], true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# bits3 ≤ # bits2, ∀ bits, bits1. # bits1 < #(bits3 + false ') → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits3 = [], bits2 = [], bitsxor(bits2, bits3) = bits2, true ∈ bits3

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;