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TRS Stand 20472 pair #381712738
details
property
value
status
complete
benchmark
32.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n106.star.cs.uiowa.edu
space
Der95
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.135593891144 seconds
cpu usage
0.112927124
max memory
6152192.0
stage attributes
key
value
output-size
5769
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 choose : [o * o * o * o] --> o cons : [o * o] --> o insert : [o * o] --> o nil : [] --> o s : [o] --> o sort : [o] --> o sort(nil) => nil sort(cons(X, Y)) => insert(X, sort(Y)) insert(X, nil) => cons(X, nil) insert(X, cons(Y, Z)) => choose(X, cons(Y, Z), X, Y) choose(X, cons(Y, Z), U, 0) => cons(X, cons(Y, Z)) choose(X, cons(Y, Z), 0, s(U)) => cons(Y, insert(X, Z)) choose(X, cons(Y, Z), s(U), s(V)) => choose(X, cons(Y, Z), U, V) 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] sort#(cons(X, Y)) =#> insert#(X, sort(Y)) 1] sort#(cons(X, Y)) =#> sort#(Y) 2] insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y) 3] choose#(X, cons(Y, Z), 0, s(U)) =#> insert#(X, Z) 4] choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V) Rules R_0: sort(nil) => nil sort(cons(X, Y)) => insert(X, sort(Y)) insert(X, nil) => cons(X, nil) insert(X, cons(Y, Z)) => choose(X, cons(Y, Z), X, Y) choose(X, cons(Y, Z), U, 0) => cons(X, cons(Y, Z)) choose(X, cons(Y, Z), 0, s(U)) => cons(Y, insert(X, Z)) choose(X, cons(Y, Z), s(U), s(V)) => choose(X, cons(Y, Z), U, V) 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 : 2 * 1 : 0, 1 * 2 : 3, 4 * 3 : 2 * 4 : 3, 4 This graph has the following strongly connected components: P_1: sort#(cons(X, Y)) =#> sort#(Y) P_2: insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y) choose#(X, cons(Y, Z), 0, s(U)) =#> insert#(X, Z) choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V) 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(choose#) = 2 nu(insert#) = 2 Thus, we can orient the dependency pairs as follows: nu(insert#(X, cons(Y, Z))) = cons(Y, Z) = cons(Y, Z) = nu(choose#(X, cons(Y, Z), X, Y)) nu(choose#(X, cons(Y, Z), 0, s(U))) = cons(Y, Z) |> Z = nu(insert#(X, Z)) nu(choose#(X, cons(Y, Z), s(U), s(V))) = cons(Y, Z) = cons(Y, Z) = nu(choose#(X, cons(Y, Z), U, V)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains: insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y) choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, 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:
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