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Proof step 10 from proof bits2uint-zeros (previous step)

bits ≠ [] ∧ bits.last, ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits.butlast) * 2 + 1 = 0, true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
bits.last, ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits = [], true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 2 * i ≠ ~1;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;