TRS Standard pair #487074013

details

property	value
status	complete
benchmark	ternary.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n178.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.0471 seconds
cpu usage	4.55794
user time	4.3721
system time	0.18584
max virtual memory	1.8409712E7
max residence set size	274768.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 45 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) MRRProof [EQUIVALENT, 41 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES


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

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

   0(#) -> #
   +(#, x) -> x
   +(x, #) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(0(x), j(y)) -> j(+(x, y))
   +(j(x), 0(y)) -> j(+(x, y))
   +(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
   +(j(x), j(y)) -> 1(+(+(x, y), j(#)))
   +(1(x), j(y)) -> 0(+(x, y))
   +(j(x), 1(y)) -> 0(+(x, y))
   +(+(x, y), z) -> +(x, +(y, z))
   opp(#) -> #
   opp(0(x)) -> 0(opp(x))
   opp(1(x)) -> j(opp(x))
   opp(j(x)) -> 1(opp(x))
   -(x, y) -> +(x, opp(y))
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(j(x), y) -> -(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))

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:

   +^1(0(x), 0(y)) -> 0^1(+(x, y))
   +^1(0(x), 0(y)) -> +^1(x, y)
   +^1(0(x), 1(y)) -> +^1(x, y)
   +^1(1(x), 0(y)) -> +^1(x, y)
   +^1(0(x), j(y)) -> +^1(x, y)
   +^1(j(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(j(x), j(y)) -> +^1(+(x, y), j(#))
   +^1(j(x), j(y)) -> +^1(x, y)
   +^1(1(x), j(y)) -> 0^1(+(x, y))
   +^1(1(x), j(y)) -> +^1(x, y)
   +^1(j(x), 1(y)) -> 0^1(+(x, y))
   +^1(j(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)
   OPP(0(x)) -> 0^1(opp(x))
   OPP(0(x)) -> OPP(x)
   OPP(1(x)) -> OPP(x)
   OPP(j(x)) -> OPP(x)
   -^1(x, y) -> +^1(x, opp(y))
   -^1(x, y) -> OPP(y)
   *^1(0(x), y) -> 0^1(*(x, y))
   *^1(0(x), y) -> *^1(x, y)
   *^1(1(x), y) -> +^1(0(*(x, y)), y)
   *^1(1(x), y) -> 0^1(*(x, y))
   *^1(1(x), y) -> *^1(x, y)

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