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:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

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:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

(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:

EXP(x, s(y)) → *¹(x, exp(x, y))
EXP(x, s(y)) → EXP(x, y)
*¹(s(x), y) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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.

-¹(s(x), s(y)) → -¹(x, y)

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

Recursive Path Order [RPO].
Precedence:

s₁ > -^1₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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:

*¹(s(x), y) → *¹(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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.

*¹(s(x), y) → *¹(x, y)

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

Recursive Path Order [RPO].
Precedence:

s₁ > *^1₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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

(17) Obligation:

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

EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

EXP(x, s(y)) → EXP(x, y)

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

Recursive Path Order [RPO].
Precedence:

s₁ > EXP₁

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE