TRS Standard pair #487069208

details

property	value
status	complete
benchmark	cade03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n183.star.cs.uiowa.edu
space	GTSSK07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.23839 seconds
cpu usage	5.42745
user time	5.19888
system time	0.228576
max virtual memory	1.841096E7
max residence set size	331112.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 5 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) NonInfProof [EQUIVALENT, 28 ms]
        (20) QDP
        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
        (22) TRUE


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

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

   minus(x, y) -> cond(ge(x, s(y)), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

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:

   minus(x, y) -> cond(ge(x, s(y)), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))


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

(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:

   MINUS(x, y) -> COND(ge(x, s(y)), x, y)
   MINUS(x, y) -> GE(x, s(y))
   COND(true, x, y) -> MINUS(x, s(y))
   GE(s(u), s(v)) -> GE(u, v)

The TRS R consists of the following rules:

   minus(x, y) -> cond(ge(x, s(y)), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

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