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Proof step 109 from proof bytes2int-int2bytes-len-eq (previous step)

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) '

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 110:
i * j + j0 = i0 * j + j1, 0 ≤ j0, j0 < j, 0 ≤ j1, j1 < j ⊦ i = i0 ∧ j0 = j1
Proof step 111:
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) '

Rule argument

seq: i * j + j0 = i0 * j + j1, 0 ≤ j0, j0 < j, 0 ≤ j1, j1 < j ⊦ i = i0 ∧ j0 = j1 subst: [i, j, j0, i0, j1] → [0, 256, 0, b2i(by0) + 1, band(b2i(by), 255)] rotate: F spec: int-mult inst: name: modeqp

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ band(i, j) = band(j, i);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 ≤ i → ¬ band(i, j) < 0;% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ 256 ≤ band(i, 255); Used terms: 0, 256, 0, b2i(by0) + 1, band(b2i(by), 255)