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Proof step 182 from proof bytes2int-int2bytes-len-eq (previous step)

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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [bytes0] → [bytes1]

Additional information

Used simplifier rules: ⊦ band(i, j) = band(j, i);% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bytes1