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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 MRRProof (⇔)
    ↳9 QDP
    ↳10 Narrowing (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 QDP
    ↳14 Narrowing (⇔)
    ↳15 QDP
    ↳16 DependencyGraphProof (⇔)
    ↳17 QDP
    ↳18 Narrowing (⇔)
    ↳19 QDP
    ↳20 DependencyGraphProof (⇔)
    ↳21 QDP
      ↳22 SemLabProof (⇐)
        ↳23 QDP
        ↳24 DependencyGraphProof (⇔)
        ↳25 QDP
        ↳26 UsableRulesReductionPairsProof (⇔)
        ↳27 QDP
      ↳28 SemLabProof2 (⇔)
        ↳29 QDP
        ↳30 UsableRulesReductionPairsProof (⇔)
        ↳31 QDP
        ↳32 RFCMatchBoundsDPProof (⇔)
        ↳33 TRUE
  ↳34 QDP
    ↳35 Narrowing (⇔)
    ↳36 QDP
    ↳37 QDPOrderProof (⇔)
    ↳38 QDP
    ↳39 PisEmptyProof (⇔)
    ↳40 TRUE

(0) Obligation:

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

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

G(x, x) → G(a, b)
G(c, g(c, x)) → G(e, g(d, x))
G(c, g(c, x)) → G(d, x)
G(d, g(d, x)) → G(c, g(e, x))
G(d, g(d, x)) → G(e, x)
G(e, g(e, x)) → G(d, g(c, x))
G(e, g(e, x)) → G(c, x)
F(g(x, y)) → G(y, g(f(f(x)), a))
F(g(x, y)) → G(f(f(x)), a)
F(g(x, y)) → F(f(x))
F(g(x, y)) → F(x)

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(e, g(e, x)) → G(d, g(c, x))
G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, x)) → G(e, g(d, x))
G(e, g(e, x)) → G(c, x)
G(c, g(c, x)) → G(d, x)
G(d, g(d, x)) → G(e, x)

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

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

(6) 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.

(7) Obligation:

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

G(e, g(e, x)) → G(d, g(c, x))
G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, x)) → G(e, g(d, x))
G(e, g(e, x)) → G(c, x)
G(c, g(c, x)) → G(d, x)
G(d, g(d, x)) → G(e, x)

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(8) 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:

G(e, g(e, x)) → G(c, x)
G(c, g(c, x)) → G(d, x)
G(d, g(d, x)) → G(e, x)


Used ordering: Polynomial interpretation [POLO]:

POL(G(x1, x2)) = x1 + x2   
POL(a) = 0   
POL(b) = 0   
POL(c) = 0   
POL(d) = 0   
POL(e) = 0   
POL(g(x1, x2)) = 1 + x1 + x2   

(9) Obligation:

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

G(e, g(e, x)) → G(d, g(c, x))
G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, x)) → G(e, g(d, x))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(10) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(e, g(e, x)) → G(d, g(c, x)) at position [1] we obtained the following new rules [LPAR04]:

G(e, g(e, c)) → G(d, g(a, b))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))

(11) Obligation:

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

G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, x)) → G(e, g(d, x))
G(e, g(e, c)) → G(d, g(a, b))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) Obligation:

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

G(c, g(c, x)) → G(e, g(d, x))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))
G(d, g(d, x)) → G(c, g(e, x))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(14) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(c, g(c, x)) → G(e, g(d, x)) at position [1] we obtained the following new rules [LPAR04]:

G(c, g(c, d)) → G(e, g(a, b))
G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))

(15) Obligation:

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

G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))
G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, d)) → G(e, g(a, b))
G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(16) DependencyGraphProof (EQUIVALENT transformation)

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

(17) Obligation:

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

G(d, g(d, x)) → G(c, g(e, x))
G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(18) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(d, g(d, x)) → G(c, g(e, x)) at position [1] we obtained the following new rules [LPAR04]:

G(d, g(d, e)) → G(c, g(a, b))
G(d, g(d, g(e, x0))) → G(c, g(d, g(c, x0)))

(19) Obligation:

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

G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))
G(d, g(d, e)) → G(c, g(a, b))
G(d, g(d, g(e, x0))) → G(c, g(d, g(c, x0)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(20) DependencyGraphProof (EQUIVALENT transformation)

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

(21) Obligation:

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

G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))
G(d, g(d, g(e, x0))) → G(c, g(d, g(c, x0)))
G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(22) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 0
a: 0
g: 0
e: 0
b: 1
d: 0
G: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(23) Obligation:

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

G.0-0(e., g.0-0(e., g.0-0(c., x0))) → G.0-0(d., g.0-0(e., g.0-0(d., x0)))
G.0-0(d., g.0-0(d., g.0-0(e., x0))) → G.0-0(c., g.0-0(d., g.0-0(c., x0)))
G.0-0(d., g.0-0(d., g.0-1(e., x0))) → G.0-0(c., g.0-0(d., g.0-1(c., x0)))
G.0-0(e., g.0-0(e., g.0-1(c., x0))) → G.0-0(d., g.0-0(e., g.0-1(d., x0)))
G.0-0(c., g.0-0(c., g.0-0(d., x0))) → G.0-0(e., g.0-0(c., g.0-0(e., x0)))
G.0-0(c., g.0-0(c., g.0-1(d., x0))) → G.0-0(e., g.0-0(c., g.0-1(e., x0)))

The TRS R consists of the following rules:

g.0-0(e., g.0-0(e., x)) → g.0-0(d., g.0-0(c., x))
g.0-0(c., g.0-1(c., x)) → g.0-0(e., g.0-1(d., x))
g.0-0(x, x) → g.0-1(a., b.)
g.0-0(e., g.0-1(e., x)) → g.0-0(d., g.0-1(c., x))
g.0-0(c., g.0-0(c., x)) → g.0-0(e., g.0-0(d., x))
g.0-0(d., g.0-0(d., x)) → g.0-0(c., g.0-0(e., x))
g.1-1(x, x) → g.0-1(a., b.)
g.0-0(d., g.0-1(d., x)) → g.0-0(c., g.0-1(e., x))

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

(24) DependencyGraphProof (EQUIVALENT transformation)

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

(25) Obligation:

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

G.0-0(d., g.0-0(d., g.0-0(e., x0))) → G.0-0(c., g.0-0(d., g.0-0(c., x0)))
G.0-0(c., g.0-0(c., g.0-0(d., x0))) → G.0-0(e., g.0-0(c., g.0-0(e., x0)))
G.0-0(e., g.0-0(e., g.0-0(c., x0))) → G.0-0(d., g.0-0(e., g.0-0(d., x0)))

The TRS R consists of the following rules:

g.0-0(e., g.0-0(e., x)) → g.0-0(d., g.0-0(c., x))
g.0-0(c., g.0-1(c., x)) → g.0-0(e., g.0-1(d., x))
g.0-0(x, x) → g.0-1(a., b.)
g.0-0(e., g.0-1(e., x)) → g.0-0(d., g.0-1(c., x))
g.0-0(c., g.0-0(c., x)) → g.0-0(e., g.0-0(d., x))
g.0-0(d., g.0-0(d., x)) → g.0-0(c., g.0-0(e., x))
g.1-1(x, x) → g.0-1(a., b.)
g.0-0(d., g.0-1(d., x)) → g.0-0(c., g.0-1(e., x))

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

(26) 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.

No dependency pairs are removed.

The following rules are removed from R:

g.0-0(x, x) → g.0-1(a., b.)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(G.0-0(x1, x2)) = x1 + x2   
POL(a.) = 0   
POL(b.) = 0   
POL(c.) = 1   
POL(d.) = 1   
POL(e.) = 1   
POL(g.0-0(x1, x2)) = 1 + x1 + x2   
POL(g.0-1(x1, x2)) = x1 + x2   

(27) Obligation:

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

G.0-0(d., g.0-0(d., g.0-0(e., x0))) → G.0-0(c., g.0-0(d., g.0-0(c., x0)))
G.0-0(c., g.0-0(c., g.0-0(d., x0))) → G.0-0(e., g.0-0(c., g.0-0(e., x0)))
G.0-0(e., g.0-0(e., g.0-0(c., x0))) → G.0-0(d., g.0-0(e., g.0-0(d., x0)))

The TRS R consists of the following rules:

g.0-0(e., g.0-0(e., x)) → g.0-0(d., g.0-0(c., x))
g.0-0(c., g.0-0(c., x)) → g.0-0(e., g.0-0(d., x))
g.0-0(d., g.0-0(d., x)) → g.0-0(c., g.0-0(e., x))
g.0-0(d., g.0-1(d., x)) → g.0-0(c., g.0-1(e., x))
g.0-0(e., g.0-1(e., x)) → g.0-0(d., g.0-1(c., x))
g.0-0(c., g.0-1(c., x)) → g.0-0(e., g.0-1(d., x))

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

(28) SemLabProof2 (EQUIVALENT transformation)

As can be seen after transforming the QDP problem by semantic labelling [SEMLAB] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

(29) Obligation:

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

G(d, g(d, g(e, x0))) → G(c, g(d, g(c, x0)))
G(c, g(c, g(d, x0))) → G(e, g(c, g(e, x0)))
G(e, g(e, g(c, x0))) → G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

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

(30) UsableRulesReductionPairsProof (EQUIVALENT transformation)

First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

d1(d(e(x0))) → c1(d(c(x0)))
c1(c(d(x0))) → e1(c(e(x0)))
e1(e(c(x0))) → d1(e(d(x0)))

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

e(e(x)) → d(c(x))
d(d(x)) → c(e(x))
c(c(x)) → e(d(x))

Q is empty.

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.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(c(x1)) = 1 + x1   
POL(c1(x1)) = x1   
POL(d(x1)) = 1 + x1   
POL(d1(x1)) = x1   
POL(e(x1)) = 1 + x1   
POL(e1(x1)) = x1   

(31) Obligation:

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

d1(d(e(x0))) → c1(d(c(x0)))
c1(c(d(x0))) → e1(c(e(x0)))
e1(e(c(x0))) → d1(e(d(x0)))

The TRS R consists of the following rules:

e(e(x)) → d(c(x))
d(d(x)) → c(e(x))
c(c(x)) → e(d(x))

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

(32) RFCMatchBoundsDPProof (EQUIVALENT transformation)

Finiteness of the DP problem can be shown by a matchbound of 1.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

d1(d(e(x0))) → c1(d(c(x0)))
c1(c(d(x0))) → e1(c(e(x0)))
e1(e(c(x0))) → d1(e(d(x0)))

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

e(e(x)) → d(c(x))
d(d(x)) → c(e(x))
c(c(x)) → e(d(x))
d1(d(e(x0))) → c1(d(c(x0)))
c1(c(d(x0))) → e1(c(e(x0)))
e1(e(c(x0))) → d1(e(d(x0)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

989123, 989124, 989126, 989125, 989127, 989128, 989129, 989130, 989131, 989132, 989133

Node 989123 is start node and node 989124 is final node.

Those nodes are connect through the following edges:

  • 989123 to 989125 labelled d1_1(0)
  • 989123 to 989127 labelled e1_1(0)
  • 989123 to 989129 labelled c1_1(0)
  • 989124 to 989124 labelled #_1(0)
  • 989126 to 989124 labelled d_1(0)
  • 989126 to 989132 labelled c_1(1)
  • 989125 to 989126 labelled e_1(0)
  • 989127 to 989128 labelled c_1(0)
  • 989128 to 989124 labelled e_1(0)
  • 989128 to 989131 labelled d_1(1)
  • 989129 to 989130 labelled d_1(0)
  • 989130 to 989124 labelled c_1(0)
  • 989130 to 989133 labelled e_1(1)
  • 989131 to 989124 labelled c_1(1)
  • 989131 to 989133 labelled e_1(1)
  • 989132 to 989124 labelled e_1(1)
  • 989132 to 989131 labelled d_1(1)
  • 989133 to 989124 labelled d_1(1)
  • 989133 to 989132 labelled c_1(1)

(33) TRUE

(34) Obligation:

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

F(g(x, y)) → F(x)
F(g(x, y)) → F(f(x))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

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

(35) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(g(x, y)) → F(f(x)) at position [0] we obtained the following new rules [LPAR04]:

F(g(g(x0, x1), y1)) → F(g(x1, g(f(f(x0)), a)))

(36) Obligation:

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

F(g(x, y)) → F(x)
F(g(g(x0, x1), y1)) → F(g(x1, g(f(f(x0)), a)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(g(x, y)) → F(x)
F(g(g(x0, x1), y1)) → F(g(x1, g(f(f(x0)), a)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F(x1)) =
/0\
\2/
 +
/20\
\00/
·x1

POL(g(x1, x2)) =
/1\
\1/
 +
/11\
\00/
·x1 +
/00\
\11/
·x2

POL(f(x1)) =
/0\
\0/
 +
/01\
\30/
·x1

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

POL(e) =
/0\
\0/

POL(d) =
/0\
\0/

POL(c) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

f(g(x, y)) → g(y, g(f(f(x)), a))
g(x, x) → g(a, b)
g(e, g(e, x)) → g(d, g(c, x))
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))

(38) Obligation:

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

g(x, x) → g(a, b)
g(c, g(c, x)) → g(e, g(d, x))
g(d, g(d, x)) → g(c, g(e, x))
g(e, g(e, x)) → g(d, g(c, x))
f(g(x, y)) → g(y, g(f(f(x)), a))

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE