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

(0) Obligation:

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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:

MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

++¹(.(x, y), z) → ++¹(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2)  =  x1
.(x1, x2)  =  .(x1, x2)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MERGE(x1, x2)  =  x2
.(x1, x2)  =  .(x2)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MERGE(x1, x2)  =  x1
.(x1, x2)  =  .(x2)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE