TRS Condi 20667 pair #381733215

details

property	value
status	complete
benchmark	310.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n101.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74391889572 seconds
cpu usage	3.648594664
max memory	1.94256896E8

stage attributes

key	value
output-size	5993
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Quasi decreasingness of the given CTRS could be proven:

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


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

(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

   gcd(add(x, y), y) -> gcd(x, y)
   gcd(y, add(x, y)) -> gcd(x, y)
   gcd(x, 0) -> x
   gcd(0, x) -> x
   add(0, y) -> y
   add(s(x), y) -> s(add(x, y))

The conditional TRS C consists of the following conditional rules:

   gcd(x, y) -> gcd(y, x) <= leq(y, x) -> false


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

(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   gcd(x, y) -> U1(leq(y, x), x, y)
   U1(false, x, y) -> gcd(y, x)
   gcd(add(x, y), y) -> gcd(x, y)
   gcd(y, add(x, y)) -> gcd(x, y)
   gcd(x, 0) -> x
   gcd(0, x) -> x
   add(0, y) -> y
   add(s(x), y) -> s(add(x, y))

Q is empty.

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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD(x, y) -> U1^1(leq(y, x), x, y)
   U1^1(false, x, y) -> GCD(y, x)
   GCD(add(x, y), y) -> GCD(x, y)
   GCD(y, add(x, y)) -> GCD(x, y)
   ADD(s(x), y) -> ADD(x, y)

The TRS R consists of the following rules:

   gcd(x, y) -> U1(leq(y, x), x, y)
   U1(false, x, y) -> gcd(y, x)
   gcd(add(x, y), y) -> gcd(x, y)
   gcd(y, add(x, y)) -> gcd(x, y)
   gcd(x, 0) -> x
   gcd(0, x) -> x
   add(0, y) -> y
   add(s(x), y) -> s(add(x, y))

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