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TRS Stand 20472 pair #381716655
details
property
value
status
complete
benchmark
Ex18_Luc06_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n009.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.297855138779 seconds
cpu usage
0.29059473
max memory
5394432.0
stage attributes
key
value
output-size
10757
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. a : [] --> o active : [o] --> o f : [o] --> o g : [o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o top : [o] --> o active(f(f(a))) => mark(f(g(f(a)))) active(f(X)) => f(active(X)) f(mark(X)) => mark(f(X)) proper(f(X)) => f(proper(X)) proper(a) => ok(a) proper(g(X)) => g(proper(X)) f(ok(X)) => ok(f(X)) g(ok(X)) => ok(g(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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(f(a))) =#> f#(g(f(a))) 1] active#(f(f(a))) =#> g#(f(a)) 2] active#(f(f(a))) =#> f#(a) 3] active#(f(X)) =#> f#(active(X)) 4] active#(f(X)) =#> active#(X) 5] f#(mark(X)) =#> f#(X) 6] proper#(f(X)) =#> f#(proper(X)) 7] proper#(f(X)) =#> proper#(X) 8] proper#(g(X)) =#> g#(proper(X)) 9] proper#(g(X)) =#> proper#(X) 10] f#(ok(X)) =#> f#(X) 11] g#(ok(X)) =#> g#(X) 12] top#(mark(X)) =#> top#(proper(X)) 13] top#(mark(X)) =#> proper#(X) 14] top#(ok(X)) =#> top#(active(X)) 15] top#(ok(X)) =#> active#(X) Rules R_0: active(f(f(a))) => mark(f(g(f(a)))) active(f(X)) => f(active(X)) f(mark(X)) => mark(f(X)) proper(f(X)) => f(proper(X)) proper(a) => ok(a) proper(g(X)) => g(proper(X)) f(ok(X)) => ok(f(X)) g(ok(X)) => ok(g(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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 : 10 * 1 : 11 * 2 : * 3 : 5, 10 * 4 : 0, 1, 2, 3, 4 * 5 : 5, 10 * 6 : 5, 10 * 7 : 6, 7, 8, 9 * 8 : 11 * 9 : 6, 7, 8, 9 * 10 : 5, 10 * 11 : 11 * 12 : 12, 13, 14, 15 * 13 : 6, 7, 8, 9 * 14 : 12, 13, 14, 15 * 15 : 0, 1, 2, 3, 4 This graph has the following strongly connected components: P_1: active#(f(X)) =#> active#(X) P_2: f#(mark(X)) =#> f#(X) f#(ok(X)) =#> f#(X)
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