YES Termination w.r.t. Q proof of Mixed_TRS_while.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 8 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 NonInfProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES

(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) YES

(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) 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, x0, x1) → F(g(x0, x1), x0, s(x1)), F(t, x2, x3) → F(g(x2, x3), x2, s(x3)) 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 Pbound 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 Pbound.
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) = -1   
POL(g(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(t) = 0   

The following pairs are in P>:

F(t, x, y) → F(g(x, y), x, s(y))
The following pairs are in Pbound:

F(t, x, y) → F(g(x, y), x, s(y))
The following rules are usable:

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

(20) 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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES