Can AES decrypt with a wrong key? The title is asking the question in layman terms, but what I really mean to ask is this:

Let $x \in \lbrace 0,1 \rbrace^{128}$ (an arbitrary input block) and let $K \in \lbrace 0, 1 \rbrace^{128}$ (an arbitrary key for 128-bit AES). Let's mark AES decryption with $AES_K^{-1}$. My questions are:


Is $AES^{-1}_K$ defined for (arbitrary) input $x$? (i.e. Is AES onto $\lbrace 0, 1 \rbrace ^{128}$ for any key?)
If so, is $AES_K( AES^{-1}_K (x) ) = x$ ?


Are those questions trivial or do they require proof?
AES-128 uses the full set $\{0, 1\}^{128}$ as keyspace, and for each key the blockcipher is defined for each input block in $\{0, 1\}^{128}$. The same goes for AES-256, but it uses a 256-bit keyspace (but still a 128-bit block).

So the answer to 1 is yes.

For 2, we have this equation:

$$AES_K(AES_K^{-1}(x)) = x$$

We can decrypt both sides:

$$AES_K^{-1}(AES_K(AES_K^{-1}(x))) = AES_K^{-1}(x)$$

Because we know that $AES_K^{-1}(AES_K(a)) = a\:$ by the definition of the inverse we can cancel out an encryption/decryption cycle on the left hand side:

$$AES_K^{-1}(x) = AES_K^{-1}(x)$$

So it's true.