details

property	value
status	complete
benchmark	jwtpa1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n059.star.cs.uiowa.edu
space	Waldmann_06

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	1.17431902885 seconds
cpu usage	1.127548426
max memory	1.058816E7

stage attributes

key	value
output-size	2614
starexec-result	"YES"

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) ) Problem 1: Innermost Equivalent Processor: -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: F(f(a,x:S),y:S) -> F(x:S,f(a,f(h(a),a))) F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) Problem 1: SCC Processor: -> Pairs: F(f(a,x:S),y:S) -> F(x:S,f(a,f(h(a),a))) F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: F(f(a,x:S),y:S) -> F(x:S,f(a,f(h(a),a))) F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) ->->-> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) Problem 1: Reduction Pairs Processor: -> Pairs: F(f(a,x:S),y:S) -> F(x:S,f(a,f(h(a),a))) F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) -> Usable rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) ->Interpretation type: Simple mixed ->Coefficients: All rationals ->Dimension: 1 ->Bound: 2 ->Interpretation: [f](X1,X2) = X1.X2 + 1/2.X1 [a] = 2 [fSNonEmpty] = 0 [h](X) = 0 [F](X1,X2) = 1/2.X1.X2 + 1/2.X1 Problem 1: SCC Processor: -> Pairs: F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) ->->-> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) Problem 1: Reduction Pairs Processor: -> Pairs: F(f(a,x:S),y:S) -> F(y:S,f(x:S,f(a,f(h(a),a)))) -> Rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) -> Usable rules: f(f(a,x:S),y:S) -> f(y:S,f(x:S,f(a,f(h(a),a)))) ->Interpretation type: Linear ->Coefficients: popout

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

job log

popout

actions

all output return to TRS Standard