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Proof step 17 from proof ubits2bits (previous step)

0 ≤ n→i(# bits1), (bits1 + false ').first, ∀ bits0. # bits0 < #(bits1 + false ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits1) * 2 = 2 * ((bits2uint(bits1) + bits2uint(bitsinvert(bits1))) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
(bits1 + false ').first, ∀ bits0. # bits0 < #(bits1 + false ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits1) = (bits2uint(bits1) + bits2uint(bitsinvert(bits1))) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);