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Proof step 11 from proof bits2uint-split (previous step)

bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1) ∨ bits1 = [], ∀ bits2, bits. # bits2 < #(bits1 + false ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) * 2 = 2 ^ (n→i(# bits1) + 1) * bits2uint(bits) + bits2uint(bits1) * 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1), ∀ bits2, bits. # bits2 < #(bits1 + false ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) * 2 = 2 ^ (n→i(# bits1) + 1) * bits2uint(bits) + bits2uint(bits1) * 2, bits1 = []

Rule argument

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ # [] = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;