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TRS Stand 20472 pair #381713174
details
property
value
status
complete
benchmark
revlist.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n042.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.138243913651 seconds
cpu usage
0.112265855
max memory
4866048.0
stage attributes
key
value
output-size
6940
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 cons : [o * o] --> o nil : [] --> o rev : [o] --> o rev1 : [o * o] --> o rev2 : [o * o] --> o s : [o] --> o rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev(rev2(Y, Z)))) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> va cons : [va * sb] --> sb nil : [] --> sb rev : [sb] --> sb rev1 : [va * sb] --> va rev2 : [va * sb] --> sb s : [va] --> va 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] rev1#(X, cons(Y, Z)) =#> rev1#(Y, Z) 1] rev#(cons(X, Y)) =#> rev1#(X, Y) 2] rev#(cons(X, Y)) =#> rev2#(X, Y) 3] rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev(rev2(Y, Z)))) 4] rev2#(X, cons(Y, Z)) =#> rev#(rev2(Y, Z)) 5] rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z) Rules R_0: rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev(rev2(Y, Z)))) 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 : 0 * 1 : 0 * 2 : 3, 4, 5 * 3 : 1, 2 * 4 : 1, 2 * 5 : 3, 4, 5 This graph has the following strongly connected components: P_1: rev1#(X, cons(Y, Z)) =#> rev1#(Y, Z) P_2: rev#(cons(X, Y)) =#> rev2#(X, Y) rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev(rev2(Y, Z)))) rev2#(X, cons(Y, Z)) =#> rev#(rev2(Y, Z)) rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z) 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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). The formative rules of (P_2, R_0) are R_1 ::= rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) => rev(cons(X, rev(rev2(Y, Z)))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite.
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