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TRS Stand 20472 pair #381712147
details
property
value
status
complete
benchmark
elimdupl.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.135037899017 seconds
cpu usage
0.13242115
max memory
5599232.0
stage attributes
key
value
output-size
8804
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. 0 : [] --> o add : [o * o] --> o eq : [o * o] --> o false : [] --> o ifrm : [o * o * o] --> o nil : [] --> o purge : [o] --> o rm : [o * o] --> o s : [o] --> o true : [] --> o eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) rm(X, nil) => nil rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) ifrm(true, X, add(Y, Z)) => rm(X, Z) ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) purge(nil) => nil purge(add(X, Y)) => add(X, purge(rm(X, Y))) 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 : [] --> ad add : [ad * ad] --> ad eq : [ad * ad] --> jb false : [] --> jb ifrm : [jb * ad * ad] --> ad nil : [] --> ad purge : [ad] --> ad rm : [ad * ad] --> ad s : [ad] --> ad true : [] --> jb 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] eq#(s(X), s(Y)) =#> eq#(X, Y) 1] rm#(X, add(Y, Z)) =#> ifrm#(eq(X, Y), X, add(Y, Z)) 2] rm#(X, add(Y, Z)) =#> eq#(X, Y) 3] ifrm#(true, X, add(Y, Z)) =#> rm#(X, Z) 4] ifrm#(false, X, add(Y, Z)) =#> rm#(X, Z) 5] purge#(add(X, Y)) =#> purge#(rm(X, Y)) 6] purge#(add(X, Y)) =#> rm#(X, Y) Rules R_0: eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) rm(X, nil) => nil rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) ifrm(true, X, add(Y, Z)) => rm(X, Z) ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) purge(nil) => nil purge(add(X, Y)) => add(X, purge(rm(X, Y))) 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 : 3, 4 * 2 : 0 * 3 : 1, 2 * 4 : 1, 2 * 5 : 5, 6 * 6 : 1, 2 This graph has the following strongly connected components: P_1: eq#(s(X), s(Y)) =#> eq#(X, Y) P_2: rm#(X, add(Y, Z)) =#> ifrm#(eq(X, Y), X, add(Y, Z)) ifrm#(true, X, add(Y, Z)) =#> rm#(X, Z) ifrm#(false, X, add(Y, Z)) =#> rm#(X, Z) P_3:
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