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Proof step 108 from proof bytes2int-int2bytes-len-eq (previous step)

0 = (256 * b2i(by0) + band(b2i(by), 255)) + 256, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 ' + by ' = i2b(~1) ' + i2b(0) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 109:
0 = 256 * (b2i(by0) + 1) + band(b2i(by), 255), ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 ' + by ' = i2b(~1) ' + i2b(0) '

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);