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SRS Standard pair #487520255
details
property
value
status
complete
benchmark
dup09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n187.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.66397690773 seconds
cpu usage
14.88230602
max memory
1.893548032E9
stage attributes
key
value
output-size
6058
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 37 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 36 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 4 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 16 ms] (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(*(*(x1)))) -> *(*(1(1(x1)))) 1(1(*(*(x1)))) -> 0(0(#(#(x1)))) #(#(0(0(x1)))) -> 0(0(#(#(x1)))) #(#(1(1(x1)))) -> 1(1(#(#(x1)))) #(#($($(x1)))) -> *(*($($(x1)))) #(#(#(#(x1)))) -> #(#(x1)) #(#(*(*(x1)))) -> *(*(x1)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(#(x_1)) = 1 + x_1 POL($(x_1)) = x_1 POL(*(x_1)) = 1 + x_1 POL(0(x_1)) = x_1 POL(1(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: #(#(#(#(x1)))) -> #(#(x1)) #(#(*(*(x1)))) -> *(*(x1)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(*(*(x1)))) -> *(*(1(1(x1)))) 1(1(*(*(x1)))) -> 0(0(#(#(x1)))) #(#(0(0(x1)))) -> 0(0(#(#(x1)))) #(#(1(1(x1)))) -> 1(1(#(#(x1)))) #(#($($(x1)))) -> *(*($($(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(*(*(x1)))) -> 1^1(1(x1)) 0^1(0(*(*(x1)))) -> 1^1(x1) 1^1(1(*(*(x1)))) -> 0^1(0(#(#(x1)))) 1^1(1(*(*(x1)))) -> 0^1(#(#(x1))) 1^1(1(*(*(x1)))) -> #^1(#(x1)) 1^1(1(*(*(x1)))) -> #^1(x1) #^1(#(0(0(x1)))) -> 0^1(0(#(#(x1)))) #^1(#(0(0(x1)))) -> 0^1(#(#(x1))) #^1(#(0(0(x1)))) -> #^1(#(x1)) #^1(#(0(0(x1)))) -> #^1(x1) #^1(#(1(1(x1)))) -> 1^1(1(#(#(x1)))) #^1(#(1(1(x1)))) -> 1^1(#(#(x1)))
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