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TRS Stand 20472 pair #381716870
details
property
value
status
complete
benchmark
Ex6_Luc98_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n049.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.202741861343 seconds
cpu usage
0.19944369
max memory
6676480.0
stage attributes
key
value
output-size
13833
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. 0 : [] --> o activate : [o] --> o cons : [o * o] --> o first : [o * o] --> o from : [o] --> o n!6220!6220first : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o nil : [] --> o s : [o] --> o first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(X, Y) => n!6220!6220first(X, Y) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => 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] first#(s(X), cons(Y, Z)) =#> activate#(Z) 1] activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) 2] activate#(n!6220!6220first(X, Y)) =#> activate#(X) 3] activate#(n!6220!6220first(X, Y)) =#> activate#(Y) 4] activate#(n!6220!6220from(X)) =#> from#(activate(X)) 5] activate#(n!6220!6220from(X)) =#> activate#(X) 6] activate#(n!6220!6220s(X)) =#> s#(activate(X)) 7] activate#(n!6220!6220s(X)) =#> activate#(X) Rules R_0: first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(X, Y) => n!6220!6220first(X, Y) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => 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 : 1, 2, 3, 4, 5, 6, 7 * 1 : 0 * 2 : 1, 2, 3, 4, 5, 6, 7 * 3 : 1, 2, 3, 4, 5, 6, 7 * 4 : * 5 : 1, 2, 3, 4, 5, 6, 7 * 6 : * 7 : 1, 2, 3, 4, 5, 6, 7 This graph has the following strongly connected components: P_1: first#(s(X), cons(Y, Z)) =#> activate#(Z) activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) activate#(n!6220!6220first(X, Y)) =#> activate#(X) activate#(n!6220!6220first(X, Y)) =#> activate#(Y) activate#(n!6220!6220from(X)) =#> activate#(X) activate#(n!6220!6220s(X)) =#> activate#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). The formative rules of (P_1, R_0) are R_1 ::= first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(X, Y) => n!6220!6220first(X, Y) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X)
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