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TRS Stand 20472 pair #381711271
details
property
value
status
complete
benchmark
17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
Der95
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.0193538665771 seconds
cpu usage
0.015437186
max memory
1855488.0
stage attributes
key
value
output-size
3062
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x y z) (RULES .(.(x,y),z) -> .(x,.(y,z)) .(i(x),x) -> 1 .(i(y),.(y,z)) -> z .(1,x) -> x .(x,i(x)) -> 1 .(x,1) -> x .(y,.(i(y),z)) -> z i(.(x,y)) -> .(i(y),i(x)) i(i(x)) -> x i(1) -> 1 ) Problem 1: Dependency Pairs Processor: -> Pairs: .#(.(x,y),z) -> .#(x,.(y,z)) .#(.(x,y),z) -> .#(y,z) I(.(x,y)) -> .#(i(y),i(x)) I(.(x,y)) -> I(x) I(.(x,y)) -> I(y) -> Rules: .(.(x,y),z) -> .(x,.(y,z)) .(i(x),x) -> 1 .(i(y),.(y,z)) -> z .(1,x) -> x .(x,i(x)) -> 1 .(x,1) -> x .(y,.(i(y),z)) -> z i(.(x,y)) -> .(i(y),i(x)) i(i(x)) -> x i(1) -> 1 Problem 1: SCC Processor: -> Pairs: .#(.(x,y),z) -> .#(x,.(y,z)) .#(.(x,y),z) -> .#(y,z) I(.(x,y)) -> .#(i(y),i(x)) I(.(x,y)) -> I(x) I(.(x,y)) -> I(y) -> Rules: .(.(x,y),z) -> .(x,.(y,z)) .(i(x),x) -> 1 .(i(y),.(y,z)) -> z .(1,x) -> x .(x,i(x)) -> 1 .(x,1) -> x .(y,.(i(y),z)) -> z i(.(x,y)) -> .(i(y),i(x)) i(i(x)) -> x i(1) -> 1 ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: .#(.(x,y),z) -> .#(x,.(y,z)) .#(.(x,y),z) -> .#(y,z) ->->-> Rules: .(.(x,y),z) -> .(x,.(y,z)) .(i(x),x) -> 1 .(i(y),.(y,z)) -> z .(1,x) -> x .(x,i(x)) -> 1 .(x,1) -> x .(y,.(i(y),z)) -> z i(.(x,y)) -> .(i(y),i(x)) i(i(x)) -> x i(1) -> 1 ->->Cycle: ->->-> Pairs: I(.(x,y)) -> I(x) I(.(x,y)) -> I(y) ->->-> Rules: .(.(x,y),z) -> .(x,.(y,z)) .(i(x),x) -> 1 .(i(y),.(y,z)) -> z .(1,x) -> x .(x,i(x)) -> 1 .(x,1) -> x .(y,.(i(y),z)) -> z i(.(x,y)) -> .(i(y),i(x)) i(i(x)) -> x i(1) -> 1 The problem is decomposed in 2 subproblems.
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