TRS Standard pair #487073018

details

property	value
status	complete
benchmark	kabasci04.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.98861 seconds
cpu usage	4.6994
user time	4.49731
system time	0.202091
max virtual memory	1.8746876E7
max residence set size	348048.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 9 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 18 ms]
        (27) QDP
        (28) PisEmptyProof [EQUIVALENT, 0 ms]
        (29) YES


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

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

   min(x, 0) -> 0
   min(0, y) -> 0
   min(s(x), s(y)) -> s(min(x, y))
   max(x, 0) -> x
   max(0, y) -> y
   max(s(x), s(y)) -> s(max(x, y))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> s(minus(x, any(y)))
   gcd(s(x), s(y)) -> gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
   any(s(x)) -> s(s(any(x)))
   any(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:

   MIN(s(x), s(y)) -> MIN(x, y)
   MAX(s(x), s(y)) -> MAX(x, y)
   MINUS(s(x), s(y)) -> MINUS(x, any(y))
   MINUS(s(x), s(y)) -> ANY(y)
   GCD(s(x), s(y)) -> GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
   GCD(s(x), s(y)) -> MINUS(max(x, y), min(x, y))
   GCD(s(x), s(y)) -> MAX(x, y)
   GCD(s(x), s(y)) -> MIN(x, y)
   ANY(s(x)) -> ANY(x)

The TRS R consists of the following rules:

   min(x, 0) -> 0
   min(0, y) -> 0
   min(s(x), s(y)) -> s(min(x, y))
   max(x, 0) -> x
   max(0, y) -> y
   max(s(x), s(y)) -> s(max(x, y))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> s(minus(x, any(y)))
   gcd(s(x), s(y)) -> gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
   any(s(x)) -> s(s(any(x)))
   any(x) -> x

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

(3) DependencyGraphProof (EQUIVALENT)

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to TRS Standard