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Proof step 6 from proof bits2uint-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits2, bits. # bits2 < #(bits1 + boolvar ') → true → bits2uint(bits + bits2) = bits2uint(bits) * 2 ^ n→i(# bits2) + bits2uint(bits2) ∨ bits2 = [] ⊦ bits2uint(bits + bits1 + boolvar ') = bits2uint(bits1 + boolvar ') + 2 ^ (n→i(# bits1) + 1) * bits2uint(bits)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;