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Proof step 21 from proof uint2int-rest (previous step)

#(false ' + bitsinvert(bits)) ≠ 0, uint2bits(i) = false ' + true ' + bits, 2 ^ (2 + n→i(# bits)) = 1 + (bits2uint(true ') * 2 ^ n→i(#(false ' + bitsinvert(bits))) + bits2uint(false ' + bitsinvert(bits))) + i, 0 < i ⊦ 2 ^ (1 + n→i(# bits)) = 1 + bits2uint(bitsinvert(bits)) + i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
uint2bits(i) = false ' + true ' + bits, 2 ^ (2 + n→i(# bits)) = 1 + bits2uint(bitsinvert(bits)) + 2 ^ (1 + n→i(# bits)) + i, 0 < i ⊦ 2 ^ (1 + n→i(# bits)) = 1 + bits2uint(bitsinvert(bits)) + i

Rule argument

Additional information

Used simplifier rules: bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);%bits2uint-one-03: ⊦ bits2uint(true ') = 1;% ⊦ m + 1 ≠ 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ # bitsinvert(bits) = # bits;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;