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Proof step 27 from proof ubits2bits (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 28:
(bits1 + true ').first, ∀ bits0. # bits0 < #(bits1 + true ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ bits1.first, 2 ^ n→i(# bits1) = (bits2uint(bits1) + 1) + bits2uint(bitsinvert(bits1))

Rule argument

subst: [bits0] → [bits1]

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ bitsinvert([]) = [];% ⊦ # [] = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits1