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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE

(0) Obligation:

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

(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:

NORM(g(x, y)) → NORM(x)
F(x, g(y, z)) → F(x, y)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.

(6) Complex Obligation (AND)

(7) Obligation:

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

REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.


REM(g(x, y), s(z)) → REM(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REM(x1, x2)  =  x1
g(x1, x2)  =  g(x1, x2)
s(x1)  =  s
norm(x1)  =  norm(x1)
nil  =  nil
0  =  0
f(x1, x2)  =  f(x1, x2)
rem(x1, x2)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[norm1, 0] > g2 > s
f2 > nil > g2 > s

Status:
g2: [1,2]
s: multiset
norm1: [1]
nil: multiset
0: multiset
f2: multiset


The following usable rules [FROCOS05] were oriented:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

(9) Obligation:

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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:

F(x, g(y, z)) → F(x, y)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.


F(x, g(y, z)) → F(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
g(x1, x2)  =  g(x1)
norm(x1)  =  x1
nil  =  nil
0  =  0
s(x1)  =  s
f(x1, x2)  =  f(x2)
rem(x1, x2)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
f1 > g1 > s
f1 > nil > 0

Status:
F2: multiset
g1: [1]
nil: multiset
0: multiset
s: multiset
f1: multiset


The following usable rules [FROCOS05] were oriented:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

(14) Obligation:

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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:

NORM(g(x, y)) → NORM(x)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.


NORM(g(x, y)) → NORM(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
NORM(x1)  =  x1
g(x1, x2)  =  g(x1)
norm(x1)  =  norm(x1)
nil  =  nil
0  =  0
s(x1)  =  s
f(x1, x2)  =  f(x2)
rem(x1, x2)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
norm1 > s
[nil, 0] > [g1, f1] > s

Status:
g1: [1]
norm1: [1]
nil: multiset
0: multiset
s: multiset
f1: [1]


The following usable rules [FROCOS05] were oriented:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

(19) Obligation:

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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