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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesReductionPairsProof (⇔)
        ↳14 QDP
        ↳15 Instantiation (⇔)
        ↳16 QDP
        ↳17 Instantiation (⇔)
        ↳18 QDP
        ↳19 MRRProof (⇔)
        ↳20 QDP
        ↳21 Instantiation (⇔)
        ↳22 QDP
        ↳23 NonTerminationProof (⇔)
        ↳24 NO

(0) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x1, x2, x3)) = x1 + x2 + 2·x3   
POL(g(x1, x2)) = 2 + x1 + x2   
POL(h(x1, x2)) = 2·x1 + 2·x2   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

g(0, Y) → 0


(2) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

Q is empty.

(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:

H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))
G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(8) 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.

(9) Obligation:

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

G(X, s(Y)) → G(X, Y)

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

(10) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • G(X, s(Y)) → G(X, Y)
    The graph contains the following edges 1 >= 1, 2 > 2

(11) YES

(12) Obligation:

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

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(X, Z) → F(X, s(X), Z)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(13) 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:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = x1 + x2 + 2·x3   
POL(H(x1, x2)) = 2·x1 + 2·x2   
POL(g(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   

(14) Obligation:

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

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(X, Z) → F(X, s(X), Z)

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

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

(15) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule H(X, Z) → F(X, s(X), Z) we obtained the following new rules [LPAR04]:

H(0, y_2) → F(0, s(0), y_2)

(16) Obligation:

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

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(0, y_2) → F(0, s(0), y_2)

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

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

(17) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(X, Y, g(X, Y)) → H(0, g(X, Y)) we obtained the following new rules [LPAR04]:

F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

(18) Obligation:

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

H(0, y_2) → F(0, s(0), y_2)
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

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

(19) 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 rules of the TRS R:

g(X, s(Y)) → g(X, Y)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = 2·x1 + x2 + 2·x3   
POL(H(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(g(x1, x2)) = 2·x1 + 2·x2   
POL(s(x1)) = 1 + 2·x1   

(20) Obligation:

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

H(0, y_2) → F(0, s(0), y_2)
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule H(0, y_2) → F(0, s(0), y_2) we obtained the following new rules [LPAR04]:

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))

(22) Obligation:

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

F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))
H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))

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

(23) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = H(0, g(0, s(0))) evaluates to t =H(0, g(0, s(0)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))
with rule H(0, g(0, s(0))) → F(0, s(0), g(0, s(0))) at position [] and matcher [ ]

F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))
with rule F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(24) NO