YES Termination w.r.t. Q proof of Secret_06_TRS_tpa10.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPSizeChangeProof (⇔, 0 ms)
        ↳34 YES
    ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QReductionProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPSizeChangeProof (⇔, 0 ms)
        ↳41 YES
    ↳42 QDP
        ↳43 UsableRulesProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 QReductionProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 TransformationProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 TransformationProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 TransformationProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 TransformationProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 TransformationProof (⇔, 0 ms)
        ↳56 QDP
        ↳57 QDPOrderProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 PisEmptyProof (⇔, 0 ms)
        ↳60 YES

(0) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
min(0, y) → 0

The TRS R 2 is

f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The signature Sigma is {f}

(2) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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:

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
+1(s(x), y) → +1(x, y)
-1(s(x), s(y)) → -1(x, y)
*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))
F(s(x), s(y)) → -1(min(s(x), s(y)), max(s(x), s(y)))
F(s(x), s(y)) → MIN(s(x), s(y))
F(s(x), s(y)) → MAX(s(x), s(y))
F(s(x), s(y)) → *1(s(x), s(y))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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 6 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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:

-1(s(x), s(y)) → -1(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(11) Obligation:

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

-1(s(x), s(y)) → -1(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:

  • -1(s(x), s(y)) → -1(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:

+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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:

+1(s(x), y) → +1(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(18) Obligation:

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

+1(s(x), y) → +1(x, y)

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

(19) 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:

  • +1(s(x), y) → +1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) YES

(21) Obligation:

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

*1(x, s(y)) → *1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(22) 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.

(23) Obligation:

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

*1(x, s(y)) → *1(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(24) 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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(25) Obligation:

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

*1(x, s(y)) → *1(x, y)

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

(26) 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:

  • *1(x, s(y)) → *1(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

(27) YES

(28) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(29) 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.

(30) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(31) 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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(32) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

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

(33) 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:

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

(34) YES

(35) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(36) 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.

(37) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(38) 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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(39) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

(40) 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:

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

(41) YES

(42) Obligation:

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

F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(43) 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.

(44) Obligation:

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

F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(45) 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].

p(s(x0))
f(s(x0), s(x1))

(46) Obligation:

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

F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(47) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y))) at position [0,0] we obtained the following new rules [LPAR04]:

F(s(x), s(y)) → F(-(s(min(x, y)), max(s(x), s(y))), *(s(x), s(y))) → F(s(x), s(y)) → F(-(s(min(x, y)), max(s(x), s(y))), *(s(x), s(y)))

(48) Obligation:

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

F(s(x), s(y)) → F(-(s(min(x, y)), max(s(x), s(y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(49) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(s(min(x, y)), max(s(x), s(y))), *(s(x), s(y))) at position [0,1] we obtained the following new rules [LPAR04]:

F(s(x), s(y)) → F(-(s(min(x, y)), s(max(x, y))), *(s(x), s(y))) → F(s(x), s(y)) → F(-(s(min(x, y)), s(max(x, y))), *(s(x), s(y)))

(50) Obligation:

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

F(s(x), s(y)) → F(-(s(min(x, y)), s(max(x, y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(51) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(s(min(x, y)), s(max(x, y))), *(s(x), s(y))) at position [0] we obtained the following new rules [LPAR04]:

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), *(s(x), s(y))) → F(s(x), s(y)) → F(-(min(x, y), max(x, y)), *(s(x), s(y)))

(52) Obligation:

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

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), *(s(x), s(y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(53) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(x, y), max(x, y)), *(s(x), s(y))) at position [1] we obtained the following new rules [LPAR04]:

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), +(s(x), *(s(x), y))) → F(s(x), s(y)) → F(-(min(x, y), max(x, y)), +(s(x), *(s(x), y)))

(54) Obligation:

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

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), +(s(x), *(s(x), y)))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(55) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(x, y), max(x, y)), +(s(x), *(s(x), y))) at position [1] we obtained the following new rules [LPAR04]:

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), s(+(x, *(s(x), y)))) → F(s(x), s(y)) → F(-(min(x, y), max(x, y)), s(+(x, *(s(x), y))))

(56) Obligation:

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

F(s(x), s(y)) → F(-(min(x, y), max(x, y)), s(+(x, *(s(x), y))))

The TRS R consists of the following rules:

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(s(x), s(y)) → F(-(min(x, y), max(x, y)), s(+(x, *(s(x), y))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x1
s(x1)  =  s(x1)
-(x1, x2)  =  x1
min(x1, x2)  =  x1
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
0=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0
min(x, 0) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

(58) Obligation:

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

min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, s(y)) → +(x, *(x, y))
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))
max(0, y) → y
max(x, 0) → x
min(0, y) → 0
min(x, 0) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))

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

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) YES