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TRS Stand 20472 pair #381712937
details
property
value
status
complete
benchmark
4.60.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n039.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.100733995438 seconds
cpu usage
0.097180843
max memory
5685248.0
stage attributes
key
value
output-size
5046
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. !770!870 : [o * o] --> o !870 : [o * o] --> o !dot : [o * o] --> o del : [o * o] --> o if : [o * o * o] --> o min : [o * o] --> o msort : [o] --> o nil : [] --> o msort(nil) => nil msort(!dot(X, Y)) => !dot(min(X, Y), msort(del(min(X, Y), !dot(X, Y)))) min(X, nil) => X min(X, !dot(Y, Z)) => if(!770!870(X, Y), min(X, Z), min(Y, Z)) del(X, nil) => nil del(X, !dot(Y, Z)) => if(!870(X, Y), Z, !dot(Y, del(X, 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: !770!870 : [ac * ac] --> tb !870 : [ac * ac] --> tb !dot : [ac * ac] --> ac del : [ac * ac] --> ac if : [tb * ac * ac] --> ac min : [ac * ac] --> ac msort : [ac] --> ac nil : [] --> ac 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] msort#(!dot(X, Y)) =#> min#(X, Y) 1] msort#(!dot(X, Y)) =#> msort#(del(min(X, Y), !dot(X, Y))) 2] msort#(!dot(X, Y)) =#> del#(min(X, Y), !dot(X, Y)) 3] msort#(!dot(X, Y)) =#> min#(X, Y) 4] min#(X, !dot(Y, Z)) =#> min#(X, Z) 5] min#(X, !dot(Y, Z)) =#> min#(Y, Z) 6] del#(X, !dot(Y, Z)) =#> del#(X, Z) Rules R_0: msort(nil) => nil msort(!dot(X, Y)) => !dot(min(X, Y), msort(del(min(X, Y), !dot(X, Y)))) min(X, nil) => X min(X, !dot(Y, Z)) => if(!770!870(X, Y), min(X, Z), min(Y, Z)) del(X, nil) => nil del(X, !dot(Y, Z)) => if(!870(X, Y), Z, !dot(Y, del(X, 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 : 4, 5 * 1 : * 2 : 6 * 3 : 4, 5 * 4 : 4, 5 * 5 : 4, 5 * 6 : 6 This graph has the following strongly connected components: P_1: min#(X, !dot(Y, Z)) =#> min#(X, Z) min#(X, !dot(Y, Z)) =#> min#(Y, Z) P_2: del#(X, !dot(Y, Z)) =#> del#(X, 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). We apply the subterm criterion with the following projection function: nu(del#) = 2 Thus, we can orient the dependency pairs as follows: nu(del#(X, !dot(Y, Z))) = !dot(Y, Z) |> Z = nu(del#(X, Z))
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