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TRS Stand 20472 pair #381715725
details
property
value
status
complete
benchmark
Ex4_7_37_Bor03_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.18404507637 seconds
cpu usage
0.155559968
max memory
8077312.0
stage attributes
key
value
output-size
7678
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 activate : [o] --> o cons : [o * o] --> o from : [o] --> o minus : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220zWquot : [o * o] --> o nil : [] --> o quot : [o * o] --> o s : [o] --> o sel : [o * o] --> o zWquot : [o * o] --> o from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) minus(X, 0) => 0 minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) zWquot(X, nil) => nil zWquot(nil, X) => nil zWquot(cons(X, Y), cons(Z, U)) => cons(quot(X, Z), n!6220!6220zWquot(activate(Y), activate(U))) from(X) => n!6220!6220from(X) zWquot(X, Y) => n!6220!6220zWquot(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220zWquot(X, Y)) => zWquot(X, 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] sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) 1] sel#(s(X), cons(Y, Z)) =#> activate#(Z) 2] minus#(s(X), s(Y)) =#> minus#(X, Y) 3] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 4] quot#(s(X), s(Y)) =#> minus#(X, Y) 5] zWquot#(cons(X, Y), cons(Z, U)) =#> quot#(X, Z) 6] zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(Y) 7] zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(U) 8] activate#(n!6220!6220from(X)) =#> from#(X) 9] activate#(n!6220!6220zWquot(X, Y)) =#> zWquot#(X, Y) Rules R_0: from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) minus(X, 0) => 0 minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) zWquot(X, nil) => nil zWquot(nil, X) => nil zWquot(cons(X, Y), cons(Z, U)) => cons(quot(X, Z), n!6220!6220zWquot(activate(Y), activate(U))) from(X) => n!6220!6220from(X) zWquot(X, Y) => n!6220!6220zWquot(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220zWquot(X, Y)) => zWquot(X, 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 : 0, 1 * 1 : 8, 9 * 2 : 2 * 3 : * 4 : 2 * 5 : 3, 4 * 6 : 8, 9 * 7 : 8, 9 * 8 : * 9 : 5, 6, 7 This graph has the following strongly connected components: P_1: sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) P_2:
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