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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) MNOCProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) PisEmptyProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) UsableRulesReductionPairsProof [EQUIVALENT, 16 ms]
        (28) QDP
        (29) MRRProof [EQUIVALENT, 17 ms]
        (30) QDP
        (31) MRRProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) DependencyGraphProof [EQUIVALENT, 0 ms]
        (34) TRUE
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) MNOCProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (41) YES
    (42) QDP
        (43) QDPOrderProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) PisEmptyProof [EQUIVALENT, 0 ms]
        (46) YES


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

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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
   QUOT(s(x), s(y)) -> MINUS(x, y)
   PLUS(s(x), y) -> PLUS(x, y)
   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(minus(x, y), z) -> PLUS(y, z)
   APP(cons(x, l), k) -> APP(l, k)
   SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
   SUM(cons(x, cons(y, l))) -> PLUS(x, y)
   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
   SUM(app(l, cons(x, cons(y, k)))) -> APP(l, sum(cons(x, cons(y, k))))
   SUM(app(l, cons(x, cons(y, k)))) -> SUM(cons(x, cons(y, k)))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 5 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(cons(x, l), k) -> APP(l, k)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(cons(x, l), k) -> APP(l, k)

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

(8) 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:
*APP(cons(x, l), k) -> APP(l, k)
The graph contains the following edges 1 > 1, 2 >= 2


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

(9)
YES

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(13) 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:
*PLUS(s(x), y) -> PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(14)
YES

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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(16) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))

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

(18) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))

The set Q consists of the following terms:

   plus(0, x0)
   plus(s(x0), x1)

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

(20) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
The following rules are removed from R:

   plus(0, y) -> y
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(SUM(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1


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

(21)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   plus(0, x0)
   plus(s(x0), x1)

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

(22) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(23)
YES

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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(25) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   sum(cons(x, nil)) -> cons(x, nil)

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

(27) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   plus(0, y) -> y
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(SUM(x_1)) = x_1
   POL(app(x_1, x_2)) = 2*x_1 + x_2
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(nil) = 0
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1
   POL(sum(x_1)) = x_1


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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   plus(s(x), y) -> s(plus(x, y))
   sum(cons(x, nil)) -> cons(x, nil)

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

(29) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   app(nil, k) -> k
   app(l, nil) -> l

Used ordering: Polynomial interpretation [POLO]:

   POL(SUM(x_1)) = 2*x_1
   POL(app(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(nil) = 2
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1
   POL(sum(x_1)) = x_1


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

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   app(cons(x, l), k) -> cons(x, app(l, k))
   plus(s(x), y) -> s(plus(x, y))
   sum(cons(x, nil)) -> cons(x, nil)

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

(31) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   app(cons(x, l), k) -> cons(x, app(l, k))

Used ordering: Polynomial interpretation [POLO]:

   POL(SUM(x_1)) = 2*x_1
   POL(app(x_1, x_2)) = 2*x_1 + x_2
   POL(cons(x_1, x_2)) = 2 + x_1 + x_2
   POL(nil) = 0
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1
   POL(sum(x_1)) = x_1


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

   plus(s(x), y) -> s(plus(x, y))
   sum(cons(x, nil)) -> cons(x, nil)

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

(33) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(34)
TRUE

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

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

   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(36) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))

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

(38) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

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

   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))

The set Q consists of the following terms:

   plus(0, x0)
   plus(s(x0), x1)

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

(40) 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:
*MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
The graph contains the following edges 1 > 1


*MINUS(s(x), s(y)) -> MINUS(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:

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

(43) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
QUOT(x1, x2)  =  x1

s(x1)  =  s(x1)

minus(x1, x2)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   dummyConstant=1

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

   minus(x, 0) -> x
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(s(x), s(y)) -> minus(x, y)


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

(44)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
   sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(45) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(46)
YES