YES

We show the termination of R/S, where R is

  T(I(x),y) -> T(x,y)

and S is:

  T(x,y) -> T(x,I(y))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  T#(I(x),y) -> T#(x,y)

and Q consists of the rules:

  T(I(x),y) -> T(x,y)
  T(x,y) -> T(x,I(y))
  T#(x,y) -> T#(x,I(y))

The weakly monotone algebra (N^3, >_lex) with

  T#_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)
  I_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  T_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)

strictly orients the following dependency pairs:

  T#(I(x),y) -> T#(x,y)

We remove them from the set of dependency pairs.

No dependency pair remains.