It seems to me like the Grue Paradox can be rewritten as follows:

Definitions of terms:
---------------------
E = Emeralds
P = properties of emeralds
C = observed color of emeralds
G = green
B = blue
U = grue

O = the time emeralds are observed
1 = before the year 2000
2 = after the year 2000

T = current time

----------------------

Condition A:
(C == G) and (O == 1)

Condition B:
(C == B) and (not (O == 1))

Condition C:
(P == U) and (O == 1)

Condition D:
(P == U) and (not (O == 1))

Axiom A:
if ((Condition A) or (Condition B)) then (P = U)

Axiom B:
if (Condition C) then (C = G)

Axiom C:
if (Condition D) then (C = B)

Theorem I (principle of induction) holds that:
if ( Xo == Y) then probably ( Xu == Y)
where:
Xo = all observed X
Xu = all unobserved X

Axiom D:
All observed emeralds, "Eo", meet Condition A.

We proceed by observing the heretofore unobserved emeralds, "Eu",
at time 2 (O = 2).  We see that the following statements must both be true:

Induction 1.  probably (C = B)
Induction 2.  probably (C = G)

Induction 1 must be true because:
In relation to the properties of Eo, "Po", Axiom A tells us
that (Po = U) when Condition A is true.  Condition A is indeed true,
as per Axiom D.  Applying Theorem I yields:
if ( Po == U ) then probably ( Pu == U )
Axiom C now applies to Eu: (C = B), as we are observing them
at (O = 2) or (not (O == 1)), and (Pu == U)

Induction 2 must be true because:
Axiom D says that all observed emeralds have been green, and,
applying Theorem I yields:
if ( Co == G ) then probably ( Cu == G )

C can not be both probably G and probably B, and thus the contradiction.