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Proof step 8 from proof int2int-rest-01 (previous step)

int2bits(j) = true ' + false ' + bits, j < 0, j < ~1 ⊦ bits2int(false ' + bits) = j + 2 ^ (n→i(#(true ' + false ' + bits)) - 1), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
int2bits(j) = true ' + false ' + bits, j < 0, j < ~1 ⊦ j = 0, bits2uint(bits) = 2 ^ (1 + n→i(# bits)) + j

Rule argument

Additional information

Used simplifier rules: bits2int-pos: ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ 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;