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)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(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)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sqr(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)) → SQR(s(x))
SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sqr(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 1 SCC with 1 less node.

(6) 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)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sqr(x0)

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

(7) 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)  =  x1
s(x1)  =  s(x1)
sum(x1)  =  sum
0  =  0
+(x1, x2)  =  x2
sqr(x1)  =  sqr(x1)
*(x1, x2)  =  *

Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > sum > 0
s₁ > sqr₁ > *

Status:

s₁: multiset
sum: []
0: multiset
sqr₁: multiset
*: multiset

The following usable rules [FROCOS05] were oriented:

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

(8) Obligation:

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

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sqr(x0)

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE