Prolog problem: Variable and instanced variable do not unify, why? And how could I do it instead?

Okay, hello everyone!

The problem domain for my Prolog problem is cryptographic protocols.

I've a Prolog program I'm trying to run in GNU Prolog. It should work...but of course it doesn't.

I'm trying to put the gist of it here:

% two people, c (the client) and a (the attacker)
% we have two public keys (asymmetric cryptographic keys, e.g.PGP)
publicKey(k_c).
publicKey(k_a).
% we have two private keys (asymmetric cryptographic keys, e.g.PGP)
privateKey(k_a-1).
privateKey(k_c-1).
% here I define the two public/private key pairs.
keyPair(k_c,k_c-1).
keyPair(k_a,k_a-1).
% just some kind of id
id(c).
id(a).
开发者_运维问答% nonces (some kind of value that's always new and is not guessable)
nonce(n_c).
nonce(n_a).
% two functions
% enc(Data, Key) encrypts Data with Key
cryptoFunction(enc(_,_)).
% sign(Data, Key) signs Data with Key (a signature)
cryptoFunction(sign(_,_)).

% a default message sent from the client to a server
init(n_c,k_c,sign([c,k_c],k_c-1)).

% Now I want to find out all combinations that can be sent without violating the rules
% The server always checks for some kind of guard (see below)

% define the message template
init(Init_1, Init_2, Init_3) :-
% define the types
nonce(Init_1),
publicKey(Init_2),
id(Init_3_1_1),
% example:
% Init_3_1_2 means init
% third parameter of init (the sign function)
% first parameter of sign function
% second part of the concatenation
publicKey(Init_3_1_2),
privateKey(Init_3_2),
% build the message
Init_3 = sign(Init_3_1,Init_3_2),
Init_3_1 = [Init_3_1_1,Init_3_1_2],
keyPair(Init_2,SignKey).
Init_3 == sign([_,Init_2],SignKey).

The last rule of the body, "Init_3 == sign([_,Init_2],SignKey)" is the guard that the server is checking.

Now when I trace with Prolog, the last part is instantiated to

sign([c,k_c],k_c-1) == sign([_281,k_c],k_c-1)

And then fails. Why doesn't _281 instantiate to c? Everything else is okay. Do I have to use Init_3_1_1 as the variable name? Or is there another way to be able to use the guard?

I hope I explained the problem well, if not, please do tell.

Unification is the built-in predicate (=)/2, not (==)/2. Example:

?- sign([c,k_c],k_c-1) = sign([_281,k_c],k_c-1).
_281 = c.