TRS Stand 20472 pair #381713271

details

property	value
status	complete
benchmark	gcd_triple.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.12347388268 seconds
cpu usage	5.893767788
max memory	3.17382656E8

stage attributes

key	value
output-size	11468
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 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) QDPOrderProof [EQUIVALENT, 132 ms]
        (22) QDP
        (23) PisEmptyProof [EQUIVALENT, 0 ms]
        (24) 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))
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   gcd(s(x), s(y), z) -> gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
   gcd(x, s(y), s(z)) -> gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
   gcd(s(x), y, s(z)) -> gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
   gcd(x, 0, 0) -> x
   gcd(0, y, 0) -> y
   gcd(0, 0, z) -> 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:

   MIN(s(x), s(y)) -> MIN(x, y)
   MAX(s(x), s(y)) -> MAX(x, y)
   -^1(s(x), s(y)) -> -^1(x, y)
   GCD(s(x), s(y), z) -> GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
   GCD(s(x), s(y), z) -> -^1(max(x, y), min(x, y))
   GCD(s(x), s(y), z) -> MAX(x, y)
   GCD(s(x), s(y), z) -> MIN(x, y)
   GCD(x, s(y), s(z)) -> GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
   GCD(x, s(y), s(z)) -> -^1(max(y, z), min(y, z))
   GCD(x, s(y), s(z)) -> MAX(y, z)
   GCD(x, s(y), s(z)) -> MIN(y, z)
   GCD(s(x), y, s(z)) -> GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
   GCD(s(x), y, s(z)) -> -^1(max(x, z), min(x, z))
   GCD(s(x), y, s(z)) -> MAX(x, z)
   GCD(s(x), y, s(z)) -> MIN(x, z)

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))
   -(x, 0) -> x

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