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TRS Stand 20472 pair #381716210
details
property
value
status
complete
benchmark
ExConc_Zan97_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n056.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0962541103363 seconds
cpu usage
0.085899994
max memory
3579904.0
stage attributes
key
value
output-size
6229
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. active : [o] --> o f : [o] --> o g : [o] --> o h : [o] --> o mark : [o] --> o active(f(X)) => mark(g(h(f(X)))) mark(f(X)) => active(f(mark(X))) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(mark(X))) f(mark(X)) => f(X) f(active(X)) => f(X) g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(X) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] active#(f(X)) =#> mark#(g(h(f(X)))) 1] active#(f(X)) =#> g#(h(f(X))) 2] active#(f(X)) =#> h#(f(X)) 3] active#(f(X)) =#> f#(X) 4] mark#(f(X)) =#> active#(f(mark(X))) 5] mark#(f(X)) =#> f#(mark(X)) 6] mark#(f(X)) =#> mark#(X) 7] mark#(g(X)) =#> active#(g(X)) 8] mark#(g(X)) =#> g#(X) 9] mark#(h(X)) =#> active#(h(mark(X))) 10] mark#(h(X)) =#> h#(mark(X)) 11] mark#(h(X)) =#> mark#(X) 12] f#(mark(X)) =#> f#(X) 13] f#(active(X)) =#> f#(X) 14] g#(mark(X)) =#> g#(X) 15] g#(active(X)) =#> g#(X) 16] h#(mark(X)) =#> h#(X) 17] h#(active(X)) =#> h#(X) Rules R_0: active(f(X)) => mark(g(h(f(X)))) mark(f(X)) => active(f(mark(X))) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(mark(X))) f(mark(X)) => f(X) f(active(X)) => f(X) g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(X) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 7, 8 * 1 : * 2 : * 3 : 12, 13 * 4 : 0, 1, 2, 3 * 5 : 12, 13 * 6 : 4, 5, 6, 7, 8, 9, 10, 11 * 7 : * 8 : 14, 15 * 9 : * 10 : 16, 17 * 11 : 4, 5, 6, 7, 8, 9, 10, 11 * 12 : 12, 13 * 13 : 12, 13 * 14 : 14, 15 * 15 : 14, 15 * 16 : 16, 17 * 17 : 16, 17 This graph has the following strongly connected components: P_1: mark#(f(X)) =#> mark#(X) mark#(h(X)) =#> mark#(X) P_2:
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