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Proof step 4 from proof bytes2int-size-val (previous step)

⊦ 2 ^ (8 * (1 + n→i(n))) = bytes2int(i2b(0) ' + i2b(~1) ' + mklist(i2b(~1), n)) + 1

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i2b(0) ' + i2b(~1) ' + mklist(i2b(~1), n) = i2b(0) ' + mklist(i2b(~1), n + 1)

Additional information

Used simplifier rules: bytes2int-011x: ⊦ bytes2int(i2b(0) ' + mklist(i2b(~1), n)) = 2 ^ (8 * n→i(n)) - 1;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0; Used formulas: i2b(0) ' + i2b(~1) ' + mklist(i2b(~1), n) = i2b(0) ' + mklist(i2b(~1), n + 1)