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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, 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:

SUM(s(x)) → +¹(sum(x), s(x))
SUM(s(x)) → SUM(x)
+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+¹(x0, x1, x2)  =  +¹(x0)

Tags:
+¹ has argument tags [1,2,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
+¹(x1, x2)  =  +¹(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

+^1₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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.

SUM(s(x)) → SUM(x)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SUM(x0, x1)  =  SUM(x1)

Tags:
SUM has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
SUM(x1)  =  SUM
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

SUM: []
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE