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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ 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

Rule argument

Additional information

Used simplifier rules: bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;