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

(0) Obligation:

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

f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The TRS R 2 is

f(t, x, y) → f(g(x, y), x, s(y))

The signature Sigma is {f}

(2) Obligation:

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

f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

(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:

F(t, x, y) → F(g(x, y), x, s(y))
F(t, x, y) → G(x, y)
G(s(x), s(y)) → G(x, y)

The TRS R consists of the following rules:

f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

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 with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

G(s(x), s(y)) → G(x, y)

The TRS R consists of the following rules:

f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

G(s(x), s(y)) → G(x, y)

R is empty.
The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

(11) Obligation:

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

G(s(x), s(y)) → G(x, y)

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

(12) 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(s(x), s(y)) → G(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

F(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

F(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

f(t, x0, x1)
g(s(x0), 0)
g(s(x0), s(x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(t, x0, x1)

(18) Obligation:

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

F(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

g(s(x0), 0)
g(s(x0), s(x1))

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(t, x, y) → F(g(x, y), x, s(y)) we obtained the following new rules [LPAR04]:

F(t, z0, s(z1)) → F(g(z0, s(z1)), z0, s(s(z1)))

(20) Obligation:

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

F(t, z0, s(z1)) → F(g(z0, s(z1)), z0, s(s(z1)))

The TRS R consists of the following rules:

g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

g(s(x0), 0)
g(s(x0), s(x1))

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

(21) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair F(t, x, y) → F(g(x, y), x, s(y)) the following chains were created:

• We consider the chain F(t, x, y) → F(g(x, y), x, s(y)), F(t, x, y) → F(g(x, y), x, s(y)) which results in the following constraint:
  

  (1)    (F(g(x0, x1), x0, s(x1))=F(t, x2, x3) ⇒ F(t, x2, x3)≥F(g(x2, x3), x2, s(x3)))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (g(x0, x1)=t ⇒ F(t, x0, s(x1))≥F(g(x0, s(x1)), x0, s(s(x1))))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on g(x0, x1)=t which results in the following new constraints:
  

  (3)    (t=t ⇒ F(t, s(x4), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))

  
  

  (4)    (g(x6, x5)=t∧(g(x6, x5)=t ⇒ F(t, x6, s(x5))≥F(g(x6, s(x5)), x6, s(s(x5)))) ⇒ F(t, s(x6), s(s(x5)))≥F(g(s(x6), s(s(x5))), s(x6), s(s(s(x5)))))

  
  
  We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
  

  (5)    (F(t, s(x4), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (g(x6, x5)=t ⇒ F(t, x6, s(x5))≥F(g(x6, s(x5)), x6, s(s(x5)))) with σ = [ ] which results in the following new constraint:
  

  (6)    (F(t, x6, s(x5))≥F(g(x6, s(x5)), x6, s(s(x5))) ⇒ F(t, s(x6), s(s(x5)))≥F(g(s(x6), s(s(x5))), s(x6), s(s(s(x5)))))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• F(t, x, y) → F(g(x, y), x, s(y))
  
  • (F(t, s(x4), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))
    
  • (F(t, x6, s(x5))≥F(g(x6, s(x5)), x6, s(s(x5))) ⇒ F(t, s(x6), s(s(x5)))≥F(g(s(x6), s(s(x5))), s(x6), s(s(s(x5)))))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(F(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃   
POL(c) = -2   
POL(g(x₁, x₂)) = 1   
POL(s(x₁)) = 1 + x₁   
POL(t) = 1   

The following pairs are in P_>:

F(t, x, y) → F(g(x, y), x, s(y))

The following pairs are in P_bound:

F(t, x, y) → F(g(x, y), x, s(y))

The following rules are usable:

g(x, y) → g(s(x), s(y))
t → g(s(x), 0)

(22) Obligation:

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

g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

g(s(x0), 0)
g(s(x0), s(x1))

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