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Proof step 21 from proof ubits2bits (previous step)

bits ≠ [] ∧ bits.last, bits.first, ∀ bits0. # bits0 < # bits → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits) = (bits2uint(bits.butlast) * 2 + 1) + bits2uint(bitsinvert(bits)) + 1, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
bits.last, bits.first, ∀ bits0. # bits0 < # bits → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ bits = [], 2 ^ n→i(# bits) = 2 * (bits2uint(bits.butlast) + 1) + bits2uint(bitsinvert(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 + i = i * (j + 1);