Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
MARK1(d) -> A__D
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(g3(X1, X2, X3)) -> MARK1(X2)
A__B -> A__D
MARK1(c) -> A__C
MARK1(g3(X1, X2, X3)) -> MARK1(X1)
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
MARK1(a) -> A__A
MARK1(g3(X1, X2, X3)) -> MARK1(X3)
A__A1 -> A__F1(a__b)
A__H2(X, X) -> A__F1(a__k)
A__H2(X, X) -> MARK1(X)
MARK1(h2(X1, X2)) -> A__H2(mark1(X1), mark1(X2))
A__A -> A__D
A__A1 -> A__B
MARK1(b) -> A__B
A__A -> A__C
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)
A__A1 -> A__F1(a__a)
A__B -> A__C
MARK1(A) -> A__A1
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
MARK1(k) -> A__K
A__A1 -> A__A
MARK1(z2(X1, X2)) -> MARK1(X1)
MARK1(h2(X1, X2)) -> MARK1(X1)
A__H2(X, X) -> A__K
A__G3(d, X, X) -> A__A1
A__Z2(e, X) -> MARK1(X)
MARK1(g3(X1, X2, X3)) -> A__G3(mark1(X1), mark1(X2), mark1(X3))
MARK1(h2(X1, X2)) -> MARK1(X2)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
MARK1(d) -> A__D
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(g3(X1, X2, X3)) -> MARK1(X2)
A__B -> A__D
MARK1(c) -> A__C
MARK1(g3(X1, X2, X3)) -> MARK1(X1)
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
MARK1(a) -> A__A
MARK1(g3(X1, X2, X3)) -> MARK1(X3)
A__A1 -> A__F1(a__b)
A__H2(X, X) -> A__F1(a__k)
A__H2(X, X) -> MARK1(X)
MARK1(h2(X1, X2)) -> A__H2(mark1(X1), mark1(X2))
A__A -> A__D
A__A1 -> A__B
MARK1(b) -> A__B
A__A -> A__C
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)
A__A1 -> A__F1(a__a)
A__B -> A__C
MARK1(A) -> A__A1
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
MARK1(k) -> A__K
A__A1 -> A__A
MARK1(z2(X1, X2)) -> MARK1(X1)
MARK1(h2(X1, X2)) -> MARK1(X1)
A__H2(X, X) -> A__K
A__G3(d, X, X) -> A__A1
A__Z2(e, X) -> MARK1(X)
MARK1(g3(X1, X2, X3)) -> A__G3(mark1(X1), mark1(X2), mark1(X3))
MARK1(h2(X1, X2)) -> MARK1(X2)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 12 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
A__A1 -> A__F1(a__a)
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(g3(X1, X2, X3)) -> MARK1(X2)
MARK1(A) -> A__A1
MARK1(f1(X)) -> A__F1(mark1(X))
MARK1(g3(X1, X2, X3)) -> MARK1(X1)
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
A__F1(X) -> MARK1(X)
MARK1(g3(X1, X2, X3)) -> MARK1(X3)
A__A1 -> A__F1(a__b)
A__H2(X, X) -> A__F1(a__k)
A__H2(X, X) -> MARK1(X)
MARK1(z2(X1, X2)) -> MARK1(X1)
MARK1(h2(X1, X2)) -> MARK1(X1)
MARK1(h2(X1, X2)) -> A__H2(mark1(X1), mark1(X2))
A__G3(d, X, X) -> A__A1
A__Z2(e, X) -> MARK1(X)
MARK1(g3(X1, X2, X3)) -> A__G3(mark1(X1), mark1(X2), mark1(X3))
MARK1(h2(X1, X2)) -> MARK1(X2)
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(g3(X1, X2, X3)) -> MARK1(X2)
MARK1(A) -> A__A1
MARK1(g3(X1, X2, X3)) -> MARK1(X1)
MARK1(g3(X1, X2, X3)) -> MARK1(X3)
MARK1(h2(X1, X2)) -> MARK1(X1)
MARK1(h2(X1, X2)) -> A__H2(mark1(X1), mark1(X2))
MARK1(g3(X1, X2, X3)) -> A__G3(mark1(X1), mark1(X2), mark1(X3))
MARK1(h2(X1, X2)) -> MARK1(X2)
The remaining pairs can at least be oriented weakly.

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
A__A1 -> A__F1(a__a)
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(f1(X)) -> A__F1(mark1(X))
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
A__F1(X) -> MARK1(X)
A__A1 -> A__F1(a__b)
A__H2(X, X) -> A__F1(a__k)
A__H2(X, X) -> MARK1(X)
MARK1(z2(X1, X2)) -> MARK1(X1)
A__G3(d, X, X) -> A__A1
A__Z2(e, X) -> MARK1(X)
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(A) = 1   
POL(A__A1) = 0   
POL(A__F1(x1)) = 2·x1   
POL(A__G3(x1, x2, x3)) = 0   
POL(A__H2(x1, x2)) = x1 + x2   
POL(A__Z2(x1, x2)) = 2·x2   
POL(MARK1(x1)) = 2·x1   
POL(a) = 0   
POL(a__A) = 1   
POL(a__a) = 0   
POL(a__b) = 0   
POL(a__c) = 0   
POL(a__d) = 0   
POL(a__f1(x1)) = 2·x1   
POL(a__g3(x1, x2, x3)) = 1 + 2·x1 + x2 + x3   
POL(a__h2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(a__k) = 0   
POL(a__z2(x1, x2)) = x1 + x2   
POL(b) = 0   
POL(c) = 0   
POL(d) = 0   
POL(e) = 0   
POL(f1(x1)) = 2·x1   
POL(g3(x1, x2, x3)) = 1 + 2·x1 + x2 + x3   
POL(h2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(k) = 0   
POL(l) = 0   
POL(m) = 0   
POL(mark1(x1)) = x1   
POL(z2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

mark1(b) -> a__b
a__c -> c
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__g3(d, X, X) -> a__A
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
mark1(l) -> l
mark1(k) -> a__k
a__c -> e
mark1(d) -> a__d
a__z2(X1, X2) -> z2(X1, X2)
a__a -> a__d
a__A -> A
mark1(c) -> a__c
a__a -> a
a__g3(X1, X2, X3) -> g3(X1, X2, X3)
a__d -> m
a__k -> l
a__h2(X1, X2) -> h2(X1, X2)
a__b -> b
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(f1(X)) -> a__f1(mark1(X))
a__b -> a__d
mark1(a) -> a__a
a__f1(X) -> f1(X)
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(m) -> m
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
a__b -> a__c
a__d -> d
a__k -> k
mark1(e) -> e
a__a -> a__c
mark1(A) -> a__A



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
A__A1 -> A__F1(a__a)
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(f1(X)) -> A__F1(mark1(X))
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
A__F1(X) -> MARK1(X)
A__A1 -> A__F1(a__b)
A__H2(X, X) -> A__F1(a__k)
A__H2(X, X) -> MARK1(X)
MARK1(z2(X1, X2)) -> MARK1(X1)
A__G3(d, X, X) -> A__A1
A__Z2(e, X) -> MARK1(X)
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

A__Z2(e, X) -> MARK1(X)
MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
MARK1(z2(X1, X2)) -> MARK1(X1)
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(z2(X1, X2)) -> A__Z2(mark1(X1), X2)
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
MARK1(z2(X1, X2)) -> MARK1(X1)
A__F1(X) -> A__Z2(mark1(X), X)
MARK1(f1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

A__Z2(e, X) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(A) = 0   
POL(A__F1(x1)) = 3 + 2·x1   
POL(A__Z2(x1, x2)) = 2·x2   
POL(MARK1(x1)) = 2·x1   
POL(a) = 1   
POL(a__A) = 0   
POL(a__a) = 1   
POL(a__b) = 3   
POL(a__c) = 0   
POL(a__d) = 0   
POL(a__f1(x1)) = 2 + 3·x1   
POL(a__g3(x1, x2, x3)) = 0   
POL(a__h2(x1, x2)) = 0   
POL(a__k) = 0   
POL(a__z2(x1, x2)) = 2 + 2·x1 + x2   
POL(b) = 3   
POL(c) = 0   
POL(d) = 0   
POL(e) = 0   
POL(f1(x1)) = 2 + 3·x1   
POL(g3(x1, x2, x3)) = 0   
POL(h2(x1, x2)) = 0   
POL(k) = 0   
POL(l) = 0   
POL(m) = 0   
POL(mark1(x1)) = x1   
POL(z2(x1, x2)) = 2 + 2·x1 + x2   

The following usable rules [14] were oriented:

mark1(b) -> a__b
a__c -> c
a__c -> l
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__g3(d, X, X) -> a__A
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__k -> m
mark1(l) -> l
mark1(k) -> a__k
a__c -> e
mark1(d) -> a__d
a__z2(X1, X2) -> z2(X1, X2)
a__a -> a__d
a__A -> A
mark1(c) -> a__c
a__a -> a
a__g3(X1, X2, X3) -> g3(X1, X2, X3)
a__d -> m
a__k -> l
a__h2(X1, X2) -> h2(X1, X2)
a__b -> b
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(f1(X)) -> a__f1(mark1(X))
a__b -> a__d
mark1(a) -> a__a
a__f1(X) -> f1(X)
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(m) -> m
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
a__b -> a__c
a__d -> d
a__k -> k
mark1(e) -> e
a__a -> a__c
mark1(A) -> a__A



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                  ↳ QDP

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

A__Z2(e, X) -> MARK1(X)

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

A__H2(X, X) -> A__G3(mark1(X), mark1(X), a__f1(a__k))
A__A1 -> A__H2(a__f1(a__a), a__f1(a__b))
A__G3(d, X, X) -> A__A1

The TRS R consists of the following rules:

a__a -> a__c
a__b -> a__c
a__c -> e
a__k -> l
a__d -> m
a__a -> a__d
a__b -> a__d
a__c -> l
a__k -> m
a__A -> a__h2(a__f1(a__a), a__f1(a__b))
a__h2(X, X) -> a__g3(mark1(X), mark1(X), a__f1(a__k))
a__g3(d, X, X) -> a__A
a__f1(X) -> a__z2(mark1(X), X)
a__z2(e, X) -> mark1(X)
mark1(A) -> a__A
mark1(a) -> a__a
mark1(b) -> a__b
mark1(c) -> a__c
mark1(d) -> a__d
mark1(k) -> a__k
mark1(z2(X1, X2)) -> a__z2(mark1(X1), X2)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(h2(X1, X2)) -> a__h2(mark1(X1), mark1(X2))
mark1(g3(X1, X2, X3)) -> a__g3(mark1(X1), mark1(X2), mark1(X3))
mark1(e) -> e
mark1(l) -> l
mark1(m) -> m
a__A -> A
a__a -> a
a__b -> b
a__c -> c
a__d -> d
a__k -> k
a__z2(X1, X2) -> z2(X1, X2)
a__f1(X) -> f1(X)
a__h2(X1, X2) -> h2(X1, X2)
a__g3(X1, X2, X3) -> g3(X1, X2, X3)

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