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))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

SUM(s(x)) → SUM(x)
SUM1(s(x)) → SUM1(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

SUM1(s(x)) → SUM1(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUM1(s(x)) → SUM1(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM1(x1)  =  SUM1(x1)
s(x1)  =  s(x1)
sum(x1)  =  x1
0  =  0
+(x1, x2)  =  x2
sum1(x1)  =  sum1(x1)

Recursive path order with status [RPO].
Precedence:

sum1₁ > s₁
sum1₁ > 0

Status:

SUM1₁: [1]
s₁: multiset
0: multiset
sum1₁: multiset

The following usable rules [FROCOS05] were oriented:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

(9) 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))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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.

SUM(s(x)) → SUM(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
s(x1)  =  s(x1)
sum(x1)  =  x1
0  =  0
+(x1, x2)  =  x2
sum1(x1)  =  sum1(x1)

Recursive path order with status [RPO].
Precedence:

sum1₁ > s₁
sum1₁ > 0

Status:

SUM₁: [1]
s₁: multiset
0: multiset
sum1₁: multiset

The following usable rules [FROCOS05] were oriented:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

(14) 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))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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