YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) NonMonReductionPairProof [EQUIVALENT, 138 ms]
        (12) QDP
        (13) PisEmptyProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) NonMonReductionPairProof [EQUIVALENT, 149 ms]
        (17) QDP
        (18) PisEmptyProof [EQUIVALENT, 0 ms]
        (19) YES


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

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

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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)
   MINUS(x, s(y)) -> P(minus(x, y))
   MINUS(x, s(y)) -> MINUS(x, y)
   DIV(s(x), s(y)) -> DIV(minus(s(x), s(y)), s(y))
   DIV(s(x), s(y)) -> MINUS(s(x), s(y))
   LOG(s(s(x)), s(s(y))) -> LOG(div(minus(x, y), s(s(y))), s(s(y)))
   LOG(s(s(x)), s(s(y))) -> DIV(minus(x, y), s(s(y)))
   LOG(s(s(x)), s(s(y))) -> MINUS(x, y)

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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 3 SCCs with 4 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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:

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

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


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

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

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(11) NonMonReductionPairProof (EQUIVALENT)
Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:

Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:

   POL(0) = 0
   POL(DIV(x_1, x_2)) = x_1
   POL(div(x_1, x_2)) = x_2
   POL(log(x_1, x_2)) = 1
   POL(minus(x_1, x_2)) = max(0, x_1 - x_2)
   POL(p(x_1)) = max(0, -1 + x_1)
   POL(s(x_1)) = 1 + x_1


The following pairs can be oriented strictly and are deleted.

   DIV(s(x), s(y)) -> DIV(minus(s(x), s(y)), s(y))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:
   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))

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

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

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

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

(14)
YES

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

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

   LOG(s(s(x)), s(s(y))) -> LOG(div(minus(x, y), s(s(y))), s(s(y)))

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(16) NonMonReductionPairProof (EQUIVALENT)
Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:

Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:

   POL(0) = 0
   POL(LOG(x_1, x_2)) = x_1
   POL(div(x_1, x_2)) = x_1
   POL(log(x_1, x_2)) = 0
   POL(minus(x_1, x_2)) = max(0, x_1 - x_2)
   POL(p(x_1)) = max(0, -1 + x_1)
   POL(s(x_1)) = 1 + x_1


The following pairs can be oriented strictly and are deleted.

   LOG(s(s(x)), s(s(y))) -> LOG(div(minus(x, y), s(s(y))), s(s(y)))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:
   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))

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

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

   p(0) -> 0
   p(s(x)) -> x
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   minus(x, s(y)) -> p(minus(x, y))
   div(0, s(y)) -> 0
   div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y)))
   log(s(0), s(s(y))) -> 0
   log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

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

(19)
YES