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Proof step 23 from proof ubits2bits (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
(bits1 + true ').first, ∀ bits0. # bits0 < #(bits1 + true ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ (n→i(# bits1) + 1) = 2 * ((bits2uint(bits1) + 1) + bits2uint(bitsinvert(bits1)))

Rule argument

Additional information

Used simplifier rules: bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ bitsinvert(true ') = false ';% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];