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Proof step 10 from proof bits2uint-ones (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1] → [bits0]

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ # [] = 0;% ⊦ #(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: bits0