Termination w.r.t. Q proof of TRS/secret05cime3.trs

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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(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__H(X, X) → A__G(mark(X), mark(X), a__f(a__k))
MARK(d) → A__D
MARK(z(X1, X2)) → A__Z(mark(X1), X2)
MARK(g(X1, X2, X3)) → MARK(X2)
A__B → A__D
MARK(c) → A__C
MARK(g(X1, X2, X3)) → MARK(X1)
A__A¹ → A__H(a__f(a__a), a__f(a__b))
MARK(a) → A__A
MARK(g(X1, X2, X3)) → MARK(X3)
A__A¹ → A__F(a__b)
A__H(X, X) → A__F(a__k)
A__H(X, X) → MARK(X)
MARK(h(X1, X2)) → A__H(mark(X1), mark(X2))
A__A → A__D
A__A¹ → A__B
MARK(b) → A__B
A__A → A__C
A__F(X) → A__Z(mark(X), X)
MARK(f(X)) → MARK(X)
A__A¹ → A__F(a__a)
A__B → A__C
MARK(A) → A__A¹
MARK(f(X)) → A__F(mark(X))
A__F(X) → MARK(X)
MARK(k) → A__K
A__A¹ → A__A
MARK(z(X1, X2)) → MARK(X1)
MARK(h(X1, X2)) → MARK(X1)
A__H(X, X) → A__K
A__G(d, X, X) → A__A¹
A__Z(e, X) → MARK(X)
MARK(g(X1, X2, X3)) → A__G(mark(X1), mark(X2), mark(X3))
MARK(h(X1, X2)) → MARK(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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(X1, X2, X3)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

A__H(X, X) → A__G(mark(X), mark(X), a__f(a__k))
MARK(d) → A__D
MARK(z(X1, X2)) → A__Z(mark(X1), X2)
MARK(g(X1, X2, X3)) → MARK(X2)
A__B → A__D
MARK(c) → A__C
MARK(g(X1, X2, X3)) → MARK(X1)
A__A¹ → A__H(a__f(a__a), a__f(a__b))
MARK(a) → A__A
MARK(g(X1, X2, X3)) → MARK(X3)
A__A¹ → A__F(a__b)
A__H(X, X) → A__F(a__k)
A__H(X, X) → MARK(X)
MARK(h(X1, X2)) → A__H(mark(X1), mark(X2))
A__A → A__D
A__A¹ → A__B
MARK(b) → A__B
A__A → A__C
A__F(X) → A__Z(mark(X), X)
MARK(f(X)) → MARK(X)
A__A¹ → A__F(a__a)
A__B → A__C
MARK(A) → A__A¹
MARK(f(X)) → A__F(mark(X))
A__F(X) → MARK(X)
MARK(k) → A__K
A__A¹ → A__A
MARK(z(X1, X2)) → MARK(X1)
MARK(h(X1, X2)) → MARK(X1)
A__H(X, X) → A__K
A__G(d, X, X) → A__A¹
A__Z(e, X) → MARK(X)
MARK(g(X1, X2, X3)) → A__G(mark(X1), mark(X2), mark(X3))
MARK(h(X1, X2)) → MARK(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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__H(X, X) → A__G(mark(X), mark(X), a__f(a__k))
MARK(z(X1, X2)) → A__Z(mark(X1), X2)
MARK(d) → A__D
MARK(g(X1, X2, X3)) → MARK(X2)
A__B → A__D
MARK(c) → A__C
MARK(a) → A__A
A__A¹ → A__H(a__f(a__a), a__f(a__b))
MARK(g(X1, X2, X3)) → MARK(X1)
MARK(g(X1, X2, X3)) → MARK(X3)
A__A¹ → A__F(a__b)
A__H(X, X) → MARK(X)
A__H(X, X) → A__F(a__k)
MARK(h(X1, X2)) → A__H(mark(X1), mark(X2))
A__A → A__D
MARK(b) → A__B
A__A¹ → A__B
A__A → A__C
A__F(X) → A__Z(mark(X), X)
MARK(f(X)) → MARK(X)
A__A¹ → A__F(a__a)
MARK(A) → A__A¹
A__B → A__C
MARK(f(X)) → A__F(mark(X))
A__F(X) → MARK(X)
MARK(k) → A__K
A__A¹ → A__A
MARK(z(X1, X2)) → MARK(X1)
MARK(h(X1, X2)) → MARK(X1)
A__H(X, X) → A__K
A__G(d, X, X) → A__A¹
A__Z(e, X) → MARK(X)
MARK(g(X1, X2, X3)) → A__G(mark(X1), mark(X2), mark(X3))
MARK(h(X1, X2)) → MARK(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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

A__H(X, X) → A__G(mark(X), mark(X), a__f(a__k))
A__A¹ → A__F(a__a)
MARK(z(X1, X2)) → A__Z(mark(X1), X2)
MARK(g(X1, X2, X3)) → MARK(X2)
MARK(A) → A__A¹
MARK(f(X)) → A__F(mark(X))
MARK(g(X1, X2, X3)) → MARK(X1)
A__A¹ → A__H(a__f(a__a), a__f(a__b))
A__F(X) → MARK(X)
MARK(g(X1, X2, X3)) → MARK(X3)
A__A¹ → A__F(a__b)
A__H(X, X) → A__F(a__k)
A__H(X, X) → MARK(X)
MARK(z(X1, X2)) → MARK(X1)
MARK(h(X1, X2)) → MARK(X1)
MARK(h(X1, X2)) → A__H(mark(X1), mark(X2))
A__G(d, X, X) → A__A¹
A__Z(e, X) → MARK(X)
MARK(g(X1, X2, X3)) → A__G(mark(X1), mark(X2), mark(X3))
MARK(h(X1, X2)) → MARK(X2)
A__F(X) → A__Z(mark(X), X)
MARK(f(X)) → MARK(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__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(X1, X2, X3)

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