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Proof step 8 from proof bits2uint-zeros (previous step)

bits = bits1 + boolvar ', ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0, bits = [], true ∈ bits, boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ bits0. # bits0 < #(bits1 + false ') → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits1) = 0, true ∈ bits1

Rule argument

Additional information

Used simplifier rules: bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar ' ≠ [];% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;