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Proof step 9 from proof bits2uint-ones (previous step)

0 ≤ n→i(# bits0), ∀ bits1. # bits1 < #(bits0 + true ') → true → bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bits1) + 1 ∨ false ∈ bits1 ⊦ 2 ^ n→i(# bits0) * 2 = 2 * (bits2uint(bits0) + 1), false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bits1. # bits1 < #(bits0 + true ') → true → bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bits1) + 1 ∨ false ∈ bits1 ⊦ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1, false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);