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Proof step 18 from proof bytes2int-int2bytes-len-eq (previous step)

# bytes1 + 1 = # int2bytes(bytes2int(bytes1) * 256 + band(b2i(by), 255)), bytes2int(bytes1) * 256 + band(b2i(by), 255) ≤ 127, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 + by ' = int2bytes(bytes2int(bytes1) * 256 + band(b2i(by), 255)), bytes1 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) '
Proof step 20:
¬(~128 ≤ bytes2int(bytes1) * 256 + band(b2i(by), 255) ∧ bytes2int(bytes1) * 256 + band(b2i(by), 255) < 128), # bytes1 + 1 = # int2bytes(bytes2int(bytes1) * 256 + band(b2i(by), 255)), bytes2int(bytes1) * 256 + band(b2i(by), 255) ≤ 127, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 + by ' = int2bytes(bytes2int(bytes1) * 256 + band(b2i(by), 255)), bytes1 = []

Rule argument

name: int2bytes-byte rotate: F subst: [i] → [bytes2int(bytes1) * 256 + band(b2i(by), 255)]

Additional information

Used simplifier rules: ⊦ i2b(i * 256 + j) = i2b(j);% ⊦ i2b(band(255, b2i(by))) = by;% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ band(i, j) = band(j, i);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ # by ' = 0 +1;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bytes2int(bytes1) * 256 + band(b2i(by), 255)