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Proof step 13 from proof bits2uint-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ bits2, bits. # bits2 < #(bits1 + false ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits1 = [], (bits2uint(bits) * 2 ^ n→i(# bits1)) * 2 = 2 ^ (n→i(# bits1) + 1) * bits2uint(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;% ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0;