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

∀ bits2, bits. # bits2 < #(bits1 + false ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) * 2 = 2 ^ (n→i(# bits1) + 1) * bits2uint(bits) + bits2uint(bits1) * 2

User applied rule apply induction

And obtained the following new goals

Proof step 11:
bits2uint(bits + bits1) = bits2uint(bits) * 2 ^ n→i(# bits1) + bits2uint(bits1) ∨ bits1 = [], ∀ bits2, bits. # bits2 < #(bits1 + false ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1) * 2 = 2 ^ (n→i(# bits1) + 1) * bits2uint(bits) + bits2uint(bits1) * 2

Rule argument

subst: [bits2, bits] → [bits1, bits]

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, bits