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Proof step 19 from proof bits2uint-split (previous step)

0 ≤ n→i(# bits1), ∀ bits2, bits. # bits2 < #(bits1 + true ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) * 2 = bits2uint(bits1) * 2 + (2 ^ n→i(# bits1) * 2) * bits2uint(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
∀ bits2, bits. # bits2 < #(bits1 + true ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);