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

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 DependencyPairsProof (⇔)
    ↳6 QDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 AND
      ↳9 QDP
        ↳10 UsableRulesProof (⇔)
          ↳11 QDP
          ↳12 UsableRulesReductionPairsProof (⇔)
          ↳13 QDP
          ↳14 PisEmptyProof (⇔)
          ↳15 TRUE
        ↳16 UsableRulesProof (⇔)
          ↳17 QDP
      ↳18 QDP
        ↳19 UsableRulesProof (⇔)
          ↳20 QDP
          ↳21 UsableRulesReductionPairsProof (⇔)
          ↳22 QDP
          ↳23 PisEmptyProof (⇔)
          ↳24 TRUE
        ↳25 UsableRulesProof (⇔)
          ↳26 QDP
      ↳27 QDP
        ↳28 UsableRulesProof (⇔)
          ↳29 QDP
          ↳30 UsableRulesReductionPairsProof (⇔)
          ↳31 QDP
          ↳32 PisEmptyProof (⇔)
          ↳33 TRUE
        ↳34 UsableRulesProof (⇔)
          ↳35 QDP
      ↳36 QDP
        ↳37 MRRProof (⇔)
        ↳38 QDP
        ↳39 RFCMatchBoundsDPProof (⇔)
        ↳40 TRUE

(0) Obligation:

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

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

f(active(x)) → f(h(g(mark(x))))
f(mark(x)) → mark(f(active(x)))
g(mark(x)) → g(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

f(active(x)) → f(h(g(mark(x))))
f(mark(x)) → mark(f(active(x)))
g(mark(x)) → g(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

ACTIVE(f(X)) → MARK(g(h(f(X))))
ACTIVE(f(X)) → G(h(f(X)))
ACTIVE(f(X)) → H(f(X))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → H(mark(X))
MARK(h(X)) → MARK(X)
F(mark(X)) → F(X)
F(active(X)) → F(X)
G(mark(X)) → G(X)
G(active(X)) → G(X)
H(mark(X)) → H(X)
H(active(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

H(active(X)) → H(X)
H(mark(X)) → H(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(H(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(13) Obligation:

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

(18) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(19) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

G(active(X)) → G(X)
G(mark(X)) → G(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(G(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(22) Obligation:

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(26) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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

(27) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(28) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(29) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(30) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F(active(X)) → F(X)
F(mark(X)) → F(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(31) Obligation:

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(35) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(36) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(f(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(37) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(f(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(mark(x1)) = x1   

(38) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(39) RFCMatchBoundsDPProof (EQUIVALENT transformation)

Finiteness of the DP problem can be shown by a matchbound of 3.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

To find matches we regarded all rules of R and P:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

1449446, 1449447, 1449448, 1449449, 1449450, 1449451, 1449452, 1449454, 1449453, 1449455, 1449456, 1449457, 1449459, 1449458, 1449461, 1449462, 1449460, 1449463, 1449464, 1449465, 1449466, 1449467, 1449469, 1449470, 1449468

Node 1449446 is start node and node 1449447 is final node.

Those nodes are connect through the following edges:

  • 1449446 to 1449448 labelled ACTIVE_1(0), ACTIVE_1(1)
  • 1449446 to 1449447 labelled MARK_1(0), MARK_1(1)
  • 1449446 to 1449450 labelled MARK_1(0)
  • 1449446 to 1449453 labelled ACTIVE_1(0)
  • 1449446 to 1449458 labelled ACTIVE_1(1)
  • 1449446 to 1449456 labelled ACTIVE_1(1)
  • 1449446 to 1449460 labelled MARK_1(1)
  • 1449446 to 1449463 labelled MARK_1(2)
  • 1449446 to 1449466 labelled ACTIVE_1(2)
  • 1449446 to 1449467 labelled ACTIVE_1(3)
  • 1449446 to 1449468 labelled MARK_1(2)
  • 1449447 to 1449447 labelled #_1(0)
  • 1449448 to 1449449 labelled h_1(0)
  • 1449448 to 1449447 labelled g_1(0), g_1(1), h_1(1)
  • 1449448 to 1449451 labelled g_1(1)
  • 1449448 to 1449456 labelled h_1(1)
  • 1449448 to 1449448 labelled h_1(1)
  • 1449448 to 1449458 labelled h_1(1)
  • 1449448 to 1449463 labelled h_1(1)
  • 1449448 to 1449468 labelled h_1(1)
  • 1449448 to 1449467 labelled h_1(1)
  • 1449449 to 1449447 labelled mark_1(0)
  • 1449449 to 1449448 labelled active_1(1)
  • 1449449 to 1449456 labelled active_1(1)
  • 1449449 to 1449458 labelled active_1(1)
  • 1449449 to 1449463 labelled mark_1(2)
  • 1449449 to 1449467 labelled active_1(3)
  • 1449449 to 1449468 labelled mark_1(2)
  • 1449450 to 1449451 labelled g_1(0)
  • 1449451 to 1449452 labelled h_1(0)
  • 1449452 to 1449447 labelled f_1(0), f_1(1)
  • 1449454 to 1449447 labelled mark_1(0)
  • 1449454 to 1449455 labelled active_1(1)
  • 1449454 to 1449456 labelled active_1(1)
  • 1449454 to 1449458 labelled active_1(1)
  • 1449454 to 1449463 labelled mark_1(2)
  • 1449454 to 1449467 labelled active_1(3)
  • 1449454 to 1449468 labelled mark_1(2)
  • 1449453 to 1449454 labelled f_1(0)
  • 1449453 to 1449447 labelled f_1(1)
  • 1449453 to 1449455 labelled f_1(1)
  • 1449453 to 1449456 labelled f_1(1)
  • 1449453 to 1449458 labelled f_1(1)
  • 1449453 to 1449463 labelled f_1(1)
  • 1449453 to 1449468 labelled f_1(1)
  • 1449453 to 1449467 labelled f_1(1)
  • 1449455 to 1449447 labelled g_1(1)
  • 1449456 to 1449457 labelled f_1(1)
  • 1449456 to 1449447 labelled f_1(2), f_1(1)
  • 1449456 to 1449456 labelled f_1(2)
  • 1449456 to 1449458 labelled f_1(2)
  • 1449456 to 1449448 labelled f_1(2)
  • 1449456 to 1449463 labelled f_1(2)
  • 1449456 to 1449468 labelled f_1(2)
  • 1449456 to 1449467 labelled f_1(2)
  • 1449457 to 1449447 labelled mark_1(1)
  • 1449457 to 1449448 labelled active_1(1)
  • 1449457 to 1449456 labelled active_1(1)
  • 1449457 to 1449458 labelled active_1(1)
  • 1449457 to 1449468 labelled mark_1(2)
  • 1449457 to 1449463 labelled mark_1(2)
  • 1449457 to 1449467 labelled active_1(3)
  • 1449459 to 1449447 labelled mark_1(1)
  • 1449459 to 1449448 labelled active_1(1)
  • 1449459 to 1449456 labelled active_1(1)
  • 1449459 to 1449458 labelled active_1(1)
  • 1449459 to 1449468 labelled mark_1(2)
  • 1449459 to 1449463 labelled mark_1(2)
  • 1449459 to 1449467 labelled active_1(3)
  • 1449458 to 1449459 labelled h_1(1)
  • 1449458 to 1449447 labelled h_1(2), h_1(1)
  • 1449458 to 1449458 labelled h_1(2)
  • 1449458 to 1449456 labelled h_1(2)
  • 1449458 to 1449448 labelled h_1(2)
  • 1449458 to 1449463 labelled h_1(2)
  • 1449458 to 1449468 labelled h_1(2)
  • 1449458 to 1449467 labelled h_1(2)
  • 1449461 to 1449462 labelled h_1(1)
  • 1449462 to 1449454 labelled f_1(1)
  • 1449462 to 1449447 labelled f_1(1)
  • 1449462 to 1449455 labelled f_1(2), f_1(1)
  • 1449462 to 1449456 labelled f_1(2), f_1(1)
  • 1449462 to 1449458 labelled f_1(2), f_1(1)
  • 1449462 to 1449463 labelled f_1(2), f_1(1)
  • 1449462 to 1449468 labelled f_1(2), f_1(1)
  • 1449462 to 1449467 labelled f_1(2), f_1(1)
  • 1449460 to 1449461 labelled g_1(1)
  • 1449463 to 1449464 labelled g_1(2)
  • 1449464 to 1449465 labelled h_1(2)
  • 1449465 to 1449457 labelled f_1(2)
  • 1449465 to 1449456 labelled f_1(2)
  • 1449465 to 1449458 labelled f_1(2)
  • 1449465 to 1449448 labelled f_1(2)
  • 1449465 to 1449447 labelled f_1(2), f_1(1)
  • 1449465 to 1449463 labelled f_1(3), f_1(2)
  • 1449465 to 1449468 labelled f_1(3), f_1(2)
  • 1449465 to 1449467 labelled f_1(3), f_1(2)
  • 1449466 to 1449461 labelled g_1(2)
  • 1449467 to 1449464 labelled g_1(3)
  • 1449467 to 1449469 labelled g_1(3)
  • 1449469 to 1449470 labelled h_1(2)
  • 1449470 to 1449447 labelled f_1(2), f_1(1)
  • 1449468 to 1449469 labelled g_1(2)

(40) TRUE