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Proof step 3 from proof bits2uint-split (previous step)

true, ∀ bits1, bits. # bits1 < # bits0 → true → bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1) ∨ bits1 = [] ⊦ bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0) ∨ bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ bits1, bits. # bits1 < # bits0 → true → bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1) ∨ bits1 = [] ⊦ bits0 = [], bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0)

Proof step 3 from proof bits2uint-split (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information