Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 PisEmptyProof (⇔)
↳6 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(g(x, y), z) → F(y, z)

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(g(x, y), z) → F(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 0   
POL(2) = 0   
POL(F(x1, x2)) = x1   
POL(f(x1, x2)) = 1 + x1 + x2   
POL(g(x1, x2)) = 1 + x1 + x2   
POL(i(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE