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Proof step 6 from proof bits2uint-ones (previous step)

∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1 ∨ false ∈ bits0 ⊦ 2 ^ n→i(# bits) = bits2uint(bits) + 1, ∀ bits0, boolvar. bits ≠ bits0 + boolvar ', false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits1. # bits1 < #(bits0 + true ') → true → bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bits1) + 1 ∨ false ∈ bits1 ⊦ 2 ^ (n→i(# bits0) + 1) = 2 * (bits2uint(bits0) + 1), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;