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Proof step 181 from proof bytes2int-int2bytes-len-eq (previous step)

bytes = bytes1 + by ', # bytes = 1 + # int2bytes(bytes2int(bytes1)), band(b2i(by), 255) < 256, 0 < bytes2int(bytes1) * 256 + band(b2i(by), 255), 255 < bytes2int(bytes1) * 256 + band(b2i(by), 255), 0 ≤ band(b2i(by), 255), 0 ≤ bytes2int(bytes1), ∀ bytes0. # bytes0 < # bytes → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = [], # bytes = 1, bytes = int2bytes(bytes2int(bytes1)) + int2bytes(band(b2i(by), 255)).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ band(i, j) = band(j, i);% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ bytes + by ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;