TRS Standard pair #615299228

details

property	value
status	complete
benchmark	Ex1_GM03_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n024.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	1.98389983177 seconds
cpu usage	5.023041954
max memory	2.82845184E8

stage attributes

key	value
output-size	8920
starexec-result	"NO"

output

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


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


NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished


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

(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) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) MRRProof [EQUIVALENT, 5 ms]
        (16) QDP
        (17) MNOCProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) MNOCProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) NonTerminationLoopProof [COMPLETE, 0 ms]
        (22) NO


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

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

   p(0) -> 0
   p(s(X)) -> X
   leq(0, Y) -> true
   leq(s(X), 0) -> false
   leq(s(X), s(Y)) -> leq(X, Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)
   diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
   0 -> n__0
   s(X) -> n__s(X)
   activate(n__0) -> 0
   activate(n__s(X)) -> s(X)
   activate(X) -> X

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:

   LEQ(s(X), s(Y)) -> LEQ(X, Y)
   IF(true, X, Y) -> ACTIVATE(X)
   IF(false, X, Y) -> ACTIVATE(Y)
   DIFF(X, Y) -> IF(leq(X, Y), n__0, n__s(diff(p(X), Y)))
   DIFF(X, Y) -> LEQ(X, Y)
   DIFF(X, Y) -> DIFF(p(X), Y)
   DIFF(X, Y) -> P(X)
   ACTIVATE(n__0) -> 0^1
   ACTIVATE(n__s(X)) -> S(X)

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(X)) -> X
   leq(0, Y) -> true
   leq(s(X), 0) -> false
   leq(s(X), s(Y)) -> leq(X, Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)
   diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
   0 -> n__0
   s(X) -> n__s(X)
   activate(n__0) -> 0
   activate(n__s(X)) -> s(X)
   activate(X) -> X

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

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