YES Termination w.r.t. Q proof of GTSSK07_cade07.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 3 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 (⇔, 25 ms)
        ↳20 QDP
        ↳21 NonInfProof (⇔, 0 ms)
        ↳22 AND
            ↳23 QDP
                ↳24 DependencyGraphProof (⇔, 0 ms)
                ↳25 TRUE
            ↳26 QDP
                ↳27 DependencyGraphProof (⇔, 0 ms)
                ↳28 TRUE

(0) Obligation:

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

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The TRS R 2 is

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))

The signature Sigma is {f, g}

(2) Obligation:

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

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, x, y, z) → G(gt(x, y), x, y, z)
F(true, x, y, z) → GT(x, y)
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
G(true, x, y, z) → GT(x, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)

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

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

(11) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

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:

  • GT(s(u), s(v)) → GT(u, v)
    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:

G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))

The TRS R consists of the following rules:

f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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:

G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, x0, x1, x2)
g(true, x0, x1, x2)

(18) Obligation:

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

G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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 G(true, x, y, z) → F(gt(x, z), x, s(y), z) the following chains were created:
  • We consider the chain F(true, x3, x4, x5) → G(gt(x3, x4), x3, x4, x5), G(true, x6, x7, x8) → F(gt(x6, x8), x6, s(x7), x8) which results in the following constraint:
    (1)    (G(gt(x3, x4), x3, x4, x5)=G(true, x6, x7, x8) ⇒ G(true, x6, x7, x8)≥F(gt(x6, x8), x6, s(x7), x8))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x3, x4)=true ⇒ G(true, x3, x4, x5)≥F(gt(x3, x5), x3, s(x4), x5))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x3, x4)=true which results in the following new constraints:
    (3)    (true=true ⇒ G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))

    (4)    (gt(x42, x41)=true∧(∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) with σ = [x43 / x5] which results in the following new constraint:
    (6)    (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))






For Pair F(true, x, y, z) → G(gt(x, y), x, y, z) the following chains were created:
  • We consider the chain G(true, x12, x13, x14) → F(gt(x12, x14), x12, s(x13), x14), F(true, x15, x16, x17) → G(gt(x15, x16), x15, x16, x17) which results in the following constraint:
    (1)    (F(gt(x12, x14), x12, s(x13), x14)=F(true, x15, x16, x17) ⇒ F(true, x15, x16, x17)≥G(gt(x15, x16), x15, x16, x17))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x12, x14)=true ⇒ F(true, x12, s(x13), x14)≥G(gt(x12, s(x13)), x12, s(x13), x14))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x12, x14)=true which results in the following new constraints:
    (3)    (true=true ⇒ F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))

    (4)    (gt(x47, x46)=true∧(∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) with σ = [x48 / x13] which results in the following new constraint:
    (6)    (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))


  • We consider the chain G(true, x21, x22, x23) → F(gt(x21, x23), x21, x22, s(x23)), F(true, x24, x25, x26) → G(gt(x24, x25), x24, x25, x26) which results in the following constraint:
    (1)    (F(gt(x21, x23), x21, x22, s(x23))=F(true, x24, x25, x26) ⇒ F(true, x24, x25, x26)≥G(gt(x24, x25), x24, x25, x26))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x21, x23)=true ⇒ F(true, x21, x22, s(x23))≥G(gt(x21, x22), x21, x22, s(x23)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x21, x23)=true which results in the following new constraints:
    (3)    (true=true ⇒ F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))

    (4)    (gt(x52, x51)=true∧(∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) with σ = [x53 / x22] which results in the following new constraint:
    (6)    (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))






For Pair G(true, x, y, z) → F(gt(x, z), x, y, s(z)) the following chains were created:
  • We consider the chain F(true, x30, x31, x32) → G(gt(x30, x31), x30, x31, x32), G(true, x33, x34, x35) → F(gt(x33, x35), x33, x34, s(x35)) which results in the following constraint:
    (1)    (G(gt(x30, x31), x30, x31, x32)=G(true, x33, x34, x35) ⇒ G(true, x33, x34, x35)≥F(gt(x33, x35), x33, x34, s(x35)))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x30, x31)=true ⇒ G(true, x30, x31, x32)≥F(gt(x30, x32), x30, x31, s(x32)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x30, x31)=true which results in the following new constraints:
    (3)    (true=true ⇒ G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))

    (4)    (gt(x57, x56)=true∧(∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) with σ = [x58 / x32] which results in the following new constraint:
    (6)    (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))






To summarize, we get the following constraints P≥ for the following pairs.
  • G(true, x, y, z) → F(gt(x, z), x, s(y), z)
    • (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
    • (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))

  • F(true, x, y, z) → G(gt(x, y), x, y, z)
    • (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
    • (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
    • (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))
    • (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))

  • G(true, x, y, z) → F(gt(x, z), x, y, s(z))
    • (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))
    • (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))




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) = 1   
POL(F(x1, x2, x3, x4)) = x2 - x3   
POL(G(x1, x2, x3, x4)) = x2 - x3   
POL(c) = -1   
POL(false) = 1   
POL(gt(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

G(true, x, y, z) → F(gt(x, z), x, s(y), z)
The following pairs are in Pbound:

G(true, x, y, z) → F(gt(x, z), x, s(y), z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
There are no usable rules

(20) Obligation:

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

F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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(true, x, y, z) → G(gt(x, y), x, y, z) the following chains were created:
  • We consider the chain G(true, x21, x22, x23) → F(gt(x21, x23), x21, x22, s(x23)), F(true, x24, x25, x26) → G(gt(x24, x25), x24, x25, x26) which results in the following constraint:
    (1)    (F(gt(x21, x23), x21, x22, s(x23))=F(true, x24, x25, x26) ⇒ F(true, x24, x25, x26)≥G(gt(x24, x25), x24, x25, x26))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x21, x23)=true ⇒ F(true, x21, x22, s(x23))≥G(gt(x21, x22), x21, x22, s(x23)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x21, x23)=true which results in the following new constraints:
    (3)    (true=true ⇒ F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))

    (4)    (gt(x52, x51)=true∧(∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) with σ = [x53 / x22] which results in the following new constraint:
    (6)    (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))






For Pair G(true, x, y, z) → F(gt(x, z), x, y, s(z)) the following chains were created:
  • We consider the chain F(true, x30, x31, x32) → G(gt(x30, x31), x30, x31, x32), G(true, x33, x34, x35) → F(gt(x33, x35), x33, x34, s(x35)) which results in the following constraint:
    (1)    (G(gt(x30, x31), x30, x31, x32)=G(true, x33, x34, x35) ⇒ G(true, x33, x34, x35)≥F(gt(x33, x35), x33, x34, s(x35)))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gt(x30, x31)=true ⇒ G(true, x30, x31, x32)≥F(gt(x30, x32), x30, x31, s(x32)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x30, x31)=true which results in the following new constraints:
    (3)    (true=true ⇒ G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))

    (4)    (gt(x57, x56)=true∧(∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) with σ = [x58 / x32] which results in the following new constraint:
    (6)    (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))






To summarize, we get the following constraints P≥ for the following pairs.
  • F(true, x, y, z) → G(gt(x, y), x, y, z)
    • (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))
    • (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))

  • G(true, x, y, z) → F(gt(x, z), x, y, s(z))
    • (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))
    • (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))




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) = 1   
POL(F(x1, x2, x3, x4)) = x2 - x4   
POL(G(x1, x2, x3, x4)) = x2 - x4   
POL(c) = -1   
POL(false) = 1   
POL(gt(x1, x2)) = 1 + x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The following pairs are in Pbound:

F(true, x, y, z) → G(gt(x, y), x, y, z)
There are no usable rules

(22) Complex Obligation (AND)

(23) Obligation:

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

F(true, x, y, z) → G(gt(x, y), x, y, z)

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(24) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(25) TRUE

(26) Obligation:

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

G(true, x, y, z) → F(gt(x, z), x, y, s(z))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(28) TRUE