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TRS Stand 20472 pair #381713247
details
property
value
status
complete
benchmark
Ex7_BLR02_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.186173915863 seconds
cpu usage
0.182559969
max memory
6950912.0
stage attributes
key
value
output-size
10761
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 2nd : [o] --> o activate : [o] --> o cons : [o * o] --> o from : [o] --> o head : [o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o n!6220!6220take : [o * o] --> o nil : [] --> o s : [o] --> o sel : [o * o] --> o take : [o * o] --> o from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) head(cons(X, Y)) => X 2nd(cons(X, Y)) => head(activate(Y)) take(0, X) => nil take(s(X), cons(Y, Z)) => cons(Y, n!6220!6220take(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) take(X, Y) => n!6220!6220take(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220take(X, Y)) => take(activate(X), activate(Y)) 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] 2nd#(cons(X, Y)) =#> head#(activate(Y)) 1] 2nd#(cons(X, Y)) =#> activate#(Y) 2] take#(s(X), cons(Y, Z)) =#> activate#(Z) 3] sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) 4] sel#(s(X), cons(Y, Z)) =#> activate#(Z) 5] activate#(n!6220!6220from(X)) =#> from#(activate(X)) 6] activate#(n!6220!6220from(X)) =#> activate#(X) 7] activate#(n!6220!6220s(X)) =#> s#(activate(X)) 8] activate#(n!6220!6220s(X)) =#> activate#(X) 9] activate#(n!6220!6220take(X, Y)) =#> take#(activate(X), activate(Y)) 10] activate#(n!6220!6220take(X, Y)) =#> activate#(X) 11] activate#(n!6220!6220take(X, Y)) =#> activate#(Y) Rules R_0: from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) head(cons(X, Y)) => X 2nd(cons(X, Y)) => head(activate(Y)) take(0, X) => nil take(s(X), cons(Y, Z)) => cons(Y, n!6220!6220take(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) take(X, Y) => n!6220!6220take(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220take(X, Y)) => take(activate(X), activate(Y)) 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 : 5, 6, 7, 8, 9, 10, 11 * 2 : 5, 6, 7, 8, 9, 10, 11 * 3 : 3, 4 * 4 : 5, 6, 7, 8, 9, 10, 11 * 5 : * 6 : 5, 6, 7, 8, 9, 10, 11 * 7 : * 8 : 5, 6, 7, 8, 9, 10, 11 * 9 : 2 * 10 : 5, 6, 7, 8, 9, 10, 11 * 11 : 5, 6, 7, 8, 9, 10, 11 This graph has the following strongly connected components: P_1: take#(s(X), cons(Y, Z)) =#> activate#(Z)
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