TRS Stand 20472 pair #381716582

details

property	value
status	complete
benchmark	Ex49_GM04_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n039.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.49508500099 seconds
cpu usage	6.22332862
max memory	3.61824256E8

stage attributes

key	value
output-size	14194
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 1 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) QDPOrderProof [EQUIVALENT, 85 ms]
        (17) QDP
        (18) DependencyGraphProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES


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

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

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

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:

   A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)
   A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)
   A__DIV(s(X), s(Y)) -> A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   A__DIV(s(X), s(Y)) -> A__GEQ(X, Y)
   A__IF(true, X, Y) -> MARK(X)
   A__IF(false, X, Y) -> MARK(Y)
   MARK(minus(X1, X2)) -> A__MINUS(X1, X2)
   MARK(geq(X1, X2)) -> A__GEQ(X1, X2)
   MARK(div(X1, X2)) -> A__DIV(mark(X1), X2)
   MARK(div(X1, X2)) -> MARK(X1)
   MARK(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
   MARK(if(X1, X2, X3)) -> MARK(X1)
   MARK(s(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)

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