Property that prevents determining key from ciphertext and plaintext What is the property (if it is not Forward Secrecy) that prevents (good) symmetrical encryption algorithms from being vulnerable to an attack vector that requires only decrypted data, and the corresponding encrypted data, and allows the attacker to either:


Obtain the encryption key OR
Obtain some data that is useful to determine some information about the plaintext of a future intercepted ciphertext.


This is a follow-up question to Tom Leek's answer saying that the property mentioned is not Forward Secrecy, that asks, "Then what is it?"

Note that this is not about hashes, or asymmetric encryption (or any other session-token-based method), as those questions can be found at the corresponding links.

(Note, this question was moved from security.SE, which is why most of the links above lead there)
As mentioned in the comments, it appears that you are seeking to understand what provides a cipher with resistance to known plaintext attack. "Diffusion" and "Confusion" I think are the answer here.

Diffusion

"Diffusion" means that the terms/bits of the algebraic expression of the cipher are inter-dependent. Without diffusion, different subsections of the state will act independently, and so can be attacked independently. With diffusion, different subsections of the state will be dependent on the values of other subsections of the state. More diffusion leads to a greater number of terms in the expressions that represent the cipher.

Confusion

"Confusion" means that the algebraic expressions of the cipher are non-linear, which makes them difficult to work with. Non-linearity increases the degree of the ciphers equations. Ideally, the non-linear expressions are not even approximately similar to any linear expressions. 

A linear cipher can often times be easily broken with negligible time/space complexity, regardless of how many rounds or how much key material is applied. The linearity means that we can view the ciphertext as a simple equation of key + message bits, which can be easily solved if you know the message bits.