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Proof step 20 from proof bits2uint-split (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits2, bits] → [bits1, bits]

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ n < n;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # [] = 0;% ⊦ 1 * i = i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: bits1, bits