[jason1990]

Let S(t) be the stock price at time t. We'll assume there's no interest and consider calls and puts with expiration t=1 and strike price K. Denote their values at t=0 by C(K) and P(K). At expiration, the call and put are worth, respectively,

max(S(1)-K,0) and max(K-S(1),0).

Suppose you have found a CETS and a PETS. If you do the CETS and the opposite of the PETS, then at expiration you will have

max(S(1)-K,0) - max(K-S(1),0) = S(1) - K.

But you could guarantee yourself S(1)-K at time t=1 by simply borrowing K and buying the stock now for S(0). So we have

C(K) - P(K) = S(0) - K.

For this to work, we must assume that options are priced according to the startup cost of an equivalent trading strategy. In particular, we must know that such strategies exist and are unique. But we needn't assume, for example, that the stocks are performing geometric Brownian motion.

Now, you've already observed that C(S)=P(S), where S=S(0). Let a=C'(S). We might expect a<0. That is, if we increase the strike price, the value of the call will diminish. Then

C(S + dK) = C(S) + a*dK.

But P(K)=C(K)+K-S, so P'(K)=C'(K)+1, and P'(S)=a+1. Hence,

P(S - dK) = P(S) - (a+1)*dK.

In the Black-Scholes example I gave you, a=-0.3085 and we have C(S)=P(S)=22.98. This gives

C(60 + dK) = 22.98 - 0.3085*dK
P(60 - dK) = 22.98 - 0.6915*dK.

If you take dK=2 you will get very close to the values I gave you. This also shows (mathematically, though perhaps not intuitively) why the call ends up worth more than the put.