/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 12 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QReductionProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) TransformationProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) DependencyGraphProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) TransformationProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) UsableRulesProof [EQUIVALENT, 0 ms]
        (47) QDP
        (48) TransformationProof [EQUIVALENT, 0 ms]
        (49) QDP
        (50) UsableRulesProof [EQUIVALENT, 0 ms]
        (51) QDP
        (52) QReductionProof [EQUIVALENT, 0 ms]
        (53) QDP
        (54) TransformationProof [EQUIVALENT, 0 ms]
        (55) QDP
        (56) TransformationProof [EQUIVALENT, 0 ms]
        (57) QDP
        (58) TransformationProof [EQUIVALENT, 0 ms]
        (59) QDP
        (60) UsableRulesProof [EQUIVALENT, 0 ms]
        (61) QDP
        (62) TransformationProof [EQUIVALENT, 0 ms]
        (63) QDP
        (64) UsableRulesProof [EQUIVALENT, 0 ms]
        (65) QDP
        (66) QReductionProof [EQUIVALENT, 0 ms]
        (67) QDP
        (68) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (69) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

Q is empty.

----------------------------------------

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
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:

   +^1(x, s(y)) -> +^1(x, y)
   *^1(x, s(y)) -> +^1(*(x, y), x)
   *^1(x, s(y)) -> *^1(x, y)
   GE(s(x), s(y)) -> GE(x, y)
   -^1(s(x), s(y)) -> -^1(x, y)
   FACT(x) -> IFFACT(x, ge(x, s(s(0))))
   FACT(x) -> GE(x, s(s(0)))
   IFFACT(x, true) -> *^1(x, fact(-(x, s(0))))
   IFFACT(x, true) -> FACT(-(x, s(0)))
   IFFACT(x, true) -> -^1(x, s(0))

The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 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:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(8) UsableRulesProof (EQUIVALENT)
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:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(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)
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:

   GE(s(x), s(y)) -> GE(x, y)

The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(15) UsableRulesProof (EQUIVALENT)
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:

   GE(s(x), s(y)) -> GE(x, y)

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT)
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:
*GE(s(x), s(y)) -> GE(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:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(22) UsableRulesProof (EQUIVALENT)
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:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(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)
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:

   *^1(x, s(y)) -> *^1(x, y)

The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(29) UsableRulesProof (EQUIVALENT)
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:

   *^1(x, s(y)) -> *^1(x, y)

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(32)
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.
----------------------------------------

(33) QDPSizeChangeProof (EQUIVALENT)
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


----------------------------------------

(34)
YES

----------------------------------------

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(x, true) -> FACT(-(x, s(0)))
   FACT(x) -> IFFACT(x, ge(x, s(s(0))))

The TRS R consists of the following rules:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   *(x, 0) -> 0
   *(x, s(y)) -> +(*(x, y), x)
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   fact(x) -> iffact(x, ge(x, s(s(0))))
   iffact(x, true) -> *(x, fact(-(x, s(0))))
   iffact(x, false) -> s(0)

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

(36) UsableRulesProof (EQUIVALENT)
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:

   IFFACT(x, true) -> FACT(-(x, s(0)))
   FACT(x) -> IFFACT(x, ge(x, s(s(0))))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true
   -(s(x), s(y)) -> -(x, y)
   -(x, 0) -> x

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)

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

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

   +(x0, 0)
   +(x0, s(x1))
   *(x0, 0)
   *(x0, s(x1))
   fact(x0)
   iffact(x0, true)
   iffact(x0, false)


----------------------------------------

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(x, true) -> FACT(-(x, s(0)))
   FACT(x) -> IFFACT(x, ge(x, s(s(0))))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true
   -(s(x), s(y)) -> -(x, y)
   -(x, 0) -> x

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(40) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule FACT(x) -> IFFACT(x, ge(x, s(s(0)))) at position [1] we obtained the following new rules [LPAR04]:

   (FACT(0) -> IFFACT(0, false),FACT(0) -> IFFACT(0, false))
   (FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0))),FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0))))


----------------------------------------

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(x, true) -> FACT(-(x, s(0)))
   FACT(0) -> IFFACT(0, false)
   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true
   -(s(x), s(y)) -> -(x, y)
   -(x, 0) -> x

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(42) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(x, true) -> FACT(-(x, s(0)))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true
   -(s(x), s(y)) -> -(x, y)
   -(x, 0) -> x

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(44) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IFFACT(x, true) -> FACT(-(x, s(0))) at position [0] we obtained the following new rules [LPAR04]:

   (IFFACT(s(x0), true) -> FACT(-(x0, 0)),IFFACT(s(x0), true) -> FACT(-(x0, 0)))


----------------------------------------

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(x0), true) -> FACT(-(x0, 0))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true
   -(s(x), s(y)) -> -(x, y)
   -(x, 0) -> x

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(46) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(x0), true) -> FACT(-(x0, 0))

The TRS R consists of the following rules:

   -(x, 0) -> x
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(48) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IFFACT(s(x0), true) -> FACT(-(x0, 0)) at position [0] we obtained the following new rules [LPAR04]:

   (IFFACT(s(x0), true) -> FACT(x0),IFFACT(s(x0), true) -> FACT(x0))


----------------------------------------

(49)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(x0), true) -> FACT(x0)

The TRS R consists of the following rules:

   -(x, 0) -> x
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

(50) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(51)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(x0), true) -> FACT(x0)

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   -(x0, 0)
   -(s(x0), s(x1))

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

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

   -(x0, 0)
   -(s(x0), s(x1))


----------------------------------------

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(x0), true) -> FACT(x0)

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(54) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IFFACT(s(x0), true) -> FACT(x0) we obtained the following new rules [LPAR04]:

   (IFFACT(s(s(y_0)), true) -> FACT(s(y_0)),IFFACT(s(s(y_0)), true) -> FACT(s(y_0)))


----------------------------------------

(55)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0)))
   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(56) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule FACT(s(x0)) -> IFFACT(s(x0), ge(x0, s(0))) we obtained the following new rules [LPAR04]:

   (FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(s(y_0), s(0))),FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(s(y_0), s(0))))


----------------------------------------

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(s(y_0), s(0)))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(58) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(s(y_0), s(0))) at position [1] we obtained the following new rules [LPAR04]:

   (FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(y_0, 0)),FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(y_0, 0)))


----------------------------------------

(59)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(y_0, 0))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(60) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(61)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(y_0, 0))

The TRS R consists of the following rules:

   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(62) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), ge(y_0, 0)) at position [1] we obtained the following new rules [LPAR04]:

   (FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true),FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true))


----------------------------------------

(63)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true)

The TRS R consists of the following rules:

   ge(x, 0) -> true

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(64) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(65)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true)

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

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

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

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))


----------------------------------------

(67)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
   FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true)

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

(68) QDPSizeChangeProof (EQUIVALENT)
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:
*FACT(s(s(y_0))) -> IFFACT(s(s(y_0)), true)
The graph contains the following edges 1 >= 1


*IFFACT(s(s(y_0)), true) -> FACT(s(y_0))
The graph contains the following edges 1 > 1


----------------------------------------

(69)
YES