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Proof step 111 from proof bytes2int-int2bytes-len-eq (previous step)

0 = b2i(by0) + 1 ∧ 0 = band(b2i(by), 255), 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) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 112:
b2i(by) = 0, ~1 = b2i(by0), ∀ 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: ⊦ 0 ≤ i → band(i, 255) = i % 256;% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);%0 ≤ i, i < 128 ⊦ band(b2i(by), 255) = (* i) ↔ b2i(by) = i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;