TRS Stand 20472 pair #381711774

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	n111.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.71282815933 seconds
cpu usage	4.479064617
max memory	2.71867904E8

stage attributes

key	value
output-size	12762
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 6 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, 28 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))

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