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Proof step 6 from proof bytes2int-1011x (previous step)

~1 = 2 ^ (8 * (n→i(m) + 1) - 1) + bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m)) ⊦ ~1 = 2 ^ (8 * (n→i(m +1) + 1) - 1) + bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m +1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
~1 = 2 ^ (8 * (n→i(m) + 1) - 1) + bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m)) ⊦ ~1 = bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m + 1)) + 2 ^ (8 * (2 + n→i(m)) - 1)

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);% ⊦ m +1 = m + 1;