details

property	value
status	complete
benchmark	02.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Gebhardt_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	256.134232044 seconds
cpu usage	1020.86056643
max memory	1.755541504E10

stage attributes

key	value
output-size	4667
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 22 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 78 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 33.3 s] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(0(0(1(x1)))) 1(0(0(1(x1)))) -> 0(1(0(0(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(0(0(x1)))) 1(0(0(1(x1)))) -> 0(0(1(0(x1)))) 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: 0^1(0(0(0(x1)))) -> 1^1(0(0(0(x1)))) 1^1(0(0(1(x1)))) -> 0^1(0(1(0(x1)))) 1^1(0(0(1(x1)))) -> 0^1(1(0(x1))) 1^1(0(0(1(x1)))) -> 1^1(0(x1)) 1^1(0(0(1(x1)))) -> 0^1(x1) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(0(0(x1)))) 1(0(0(1(x1)))) -> 0(0(1(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(0(0(1(x1)))) -> 0^1(1(0(x1))) 1^1(0(0(1(x1)))) -> 1^1(0(x1)) 1^1(0(0(1(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( 0^1_1(x_1) ) = max{0, 2x_1 - 2} POL( 1^1_1(x_1) ) = max{0, 2x_1 - 2} POL( 1_1(x_1) ) = 2x_1 + 1 POL( 0_1(x_1) ) = 2x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 1(0(0(1(x1)))) -> 0(0(1(0(x1)))) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Stand 10685