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Proof step 9 from proof bytes2int-1011x (previous step)

m + 1 ≠ 0, ~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)-1) + i2b(~1) ') + 2 ^ (8 * (2 + n→i(m)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
~1 = 2 ^ (8 * (n→i(m) + 1) - 1) + bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m)) ⊦ ~1 = 2 ^ (8 * (2 + n→i(m)) - 1) + bytes2int(i2b(~1) ' + i2b(127) ' + mklist(i2b(~1), m) + i2b(~1) ')

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (m + 1)-1 = m;