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Proof step 12 from proof bits2uint-inv-neg (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 13:
bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bitsinvert(bits1)) + bits2uint(bits1) + 1 ∨ bits1.first, ∀ bits0. # bits0 < #(bits1 + true ') → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bitsinvert(bits0)) + bits2uint(bits0) + 1 ∨ bits0.first ⊦ 2 ^ (n→i(# bits1) + 1) = 2 * ((bits2uint(bitsinvert(bits1)) + bits2uint(bits1)) + 1), (bits1 + true ').first

Rule argument

subst: [bits0] → [bits1]

Additional information

Used simplifier rules: ⊦ #(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