TRS Standard pair #516967825

details

property	value
status	complete
benchmark	list-sum-prod-bin-assoc.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.35076904297 seconds
cpu usage	6.203283311
max memory	4.82500608E8

stage attributes

key	value
output-size	11517
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 3 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (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


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(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))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))

Q is empty.

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(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(1(x), 1(y)) -> 0^1(+(+(x, y), 1(#)))
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)
   *^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)
   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)
   SUM(nil) -> 0^1(#)
   SUM(cons(x, l)) -> +^1(x, sum(l))
   SUM(cons(x, l)) -> SUM(l)
   PROD(cons(x, l)) -> *^1(x, prod(l))
   PROD(cons(x, l)) -> PROD(l)

The TRS R consists of the following rules:

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