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TRS Stand 20472 pair #381710121
details
property
value
status
complete
benchmark
recursion-5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n105.star.cs.uiowa.edu
space
TCT_12
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.020751953125 seconds
cpu usage
0.018404313
max memory
2015232.0
stage attributes
key
value
output-size
8446
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) (RULES f_0(x) -> a f_1(x) -> g_1(x,x) f_2(x) -> g_2(x,x) f_3(x) -> g_3(x,x) f_4(x) -> g_4(x,x) f_5(x) -> g_5(x,x) g_1(s(x),y) -> b(f_0(y),g_1(x,y)) g_2(s(x),y) -> b(f_1(y),g_2(x,y)) g_3(s(x),y) -> b(f_2(y),g_3(x,y)) g_4(s(x),y) -> b(f_3(y),g_4(x,y)) g_5(s(x),y) -> b(f_4(y),g_5(x,y)) ) Problem 1: Innermost Equivalent Processor: -> Rules: f_0(x) -> a f_1(x) -> g_1(x,x) f_2(x) -> g_2(x,x) f_3(x) -> g_3(x,x) f_4(x) -> g_4(x,x) f_5(x) -> g_5(x,x) g_1(s(x),y) -> b(f_0(y),g_1(x,y)) g_2(s(x),y) -> b(f_1(y),g_2(x,y)) g_3(s(x),y) -> b(f_2(y),g_3(x,y)) g_4(s(x),y) -> b(f_3(y),g_4(x,y)) g_5(s(x),y) -> b(f_4(y),g_5(x,y)) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: F_1(x) -> G_1(x,x) F_2(x) -> G_2(x,x) F_3(x) -> G_3(x,x) F_4(x) -> G_4(x,x) F_5(x) -> G_5(x,x) G_1(s(x),y) -> F_0(y) G_1(s(x),y) -> G_1(x,y) G_2(s(x),y) -> F_1(y) G_2(s(x),y) -> G_2(x,y) G_3(s(x),y) -> F_2(y) G_3(s(x),y) -> G_3(x,y) G_4(s(x),y) -> F_3(y) G_4(s(x),y) -> G_4(x,y) G_5(s(x),y) -> F_4(y) G_5(s(x),y) -> G_5(x,y) -> Rules: f_0(x) -> a f_1(x) -> g_1(x,x) f_2(x) -> g_2(x,x) f_3(x) -> g_3(x,x) f_4(x) -> g_4(x,x) f_5(x) -> g_5(x,x) g_1(s(x),y) -> b(f_0(y),g_1(x,y)) g_2(s(x),y) -> b(f_1(y),g_2(x,y)) g_3(s(x),y) -> b(f_2(y),g_3(x,y)) g_4(s(x),y) -> b(f_3(y),g_4(x,y)) g_5(s(x),y) -> b(f_4(y),g_5(x,y)) Problem 1: SCC Processor: -> Pairs: F_1(x) -> G_1(x,x) F_2(x) -> G_2(x,x) F_3(x) -> G_3(x,x) F_4(x) -> G_4(x,x) F_5(x) -> G_5(x,x) G_1(s(x),y) -> F_0(y) G_1(s(x),y) -> G_1(x,y) G_2(s(x),y) -> F_1(y) G_2(s(x),y) -> G_2(x,y) G_3(s(x),y) -> F_2(y) G_3(s(x),y) -> G_3(x,y) G_4(s(x),y) -> F_3(y) G_4(s(x),y) -> G_4(x,y) G_5(s(x),y) -> F_4(y) G_5(s(x),y) -> G_5(x,y) -> Rules: f_0(x) -> a f_1(x) -> g_1(x,x) f_2(x) -> g_2(x,x) f_3(x) -> g_3(x,x) f_4(x) -> g_4(x,x)
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