Term Rewriting System R:
[x, y, z]
a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) -> p(a(x, z), a(y, z))
a(a(x, y), z) -> a(x, a(y, z))
a(id, x) -> x
a(1, id) -> 1
a(t, id) -> t
a(1, p(x, y)) -> x
a(t, p(x, y)) -> y

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

A(lambda(x), y) -> A(x, p(1, a(y, t)))
A(lambda(x), y) -> A(y, t)
A(p(x, y), z) -> A(x, z)
A(p(x, y), z) -> A(y, z)
A(a(x, y), z) -> A(x, a(y, z))
A(a(x, y), z) -> A(y, z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

A(a(x, y), z) -> A(y, z)
A(a(x, y), z) -> A(x, a(y, z))
A(p(x, y), z) -> A(y, z)
A(p(x, y), z) -> A(x, z)
A(lambda(x), y) -> A(y, t)
A(lambda(x), y) -> A(x, p(1, a(y, t)))


Rules:


a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) -> p(a(x, z), a(y, z))
a(a(x, y), z) -> a(x, a(y, z))
a(id, x) -> x
a(1, id) -> 1
a(t, id) -> t
a(1, p(x, y)) -> x
a(t, p(x, y)) -> y




The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes