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TRS Stand 20472 pair #381716160
details
property
value
status
complete
benchmark
Ex3_3_25_Bor03_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n006.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.544454097748 seconds
cpu usage
0.540964033
max memory
1.527808E7
stage attributes
key
value
output-size
25933
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. a!6220!6220app : [o * o] --> o a!6220!6220from : [o] --> o a!6220!6220prefix : [o] --> o a!6220!6220zWadr : [o * o] --> o app : [o * o] --> o cons : [o * o] --> o from : [o] --> o mark : [o] --> o nil : [] --> o prefix : [o] --> o s : [o] --> o zWadr : [o * o] --> o a!6220!6220app(nil, X) => mark(X) a!6220!6220app(cons(X, Y), Z) => cons(mark(X), app(Y, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220zWadr(nil, X) => nil a!6220!6220zWadr(X, nil) => nil a!6220!6220zWadr(cons(X, Y), cons(Z, U)) => cons(a!6220!6220app(mark(Z), cons(mark(X), nil)), zWadr(Y, U)) a!6220!6220prefix(X) => cons(nil, zWadr(X, prefix(X))) mark(app(X, Y)) => a!6220!6220app(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(zWadr(X, Y)) => a!6220!6220zWadr(mark(X), mark(Y)) mark(prefix(X)) => a!6220!6220prefix(mark(X)) mark(nil) => nil mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) a!6220!6220app(X, Y) => app(X, Y) a!6220!6220from(X) => from(X) a!6220!6220zWadr(X, Y) => zWadr(X, Y) a!6220!6220prefix(X) => prefix(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] a!6220!6220app#(nil, X) =#> mark#(X) 1] a!6220!6220app#(cons(X, Y), Z) =#> mark#(X) 2] a!6220!6220from#(X) =#> mark#(X) 3] a!6220!6220zWadr#(cons(X, Y), cons(Z, U)) =#> a!6220!6220app#(mark(Z), cons(mark(X), nil)) 4] a!6220!6220zWadr#(cons(X, Y), cons(Z, U)) =#> mark#(Z) 5] a!6220!6220zWadr#(cons(X, Y), cons(Z, U)) =#> mark#(X) 6] mark#(app(X, Y)) =#> a!6220!6220app#(mark(X), mark(Y)) 7] mark#(app(X, Y)) =#> mark#(X) 8] mark#(app(X, Y)) =#> mark#(Y) 9] mark#(from(X)) =#> a!6220!6220from#(mark(X)) 10] mark#(from(X)) =#> mark#(X) 11] mark#(zWadr(X, Y)) =#> a!6220!6220zWadr#(mark(X), mark(Y)) 12] mark#(zWadr(X, Y)) =#> mark#(X) 13] mark#(zWadr(X, Y)) =#> mark#(Y) 14] mark#(prefix(X)) =#> a!6220!6220prefix#(mark(X)) 15] mark#(prefix(X)) =#> mark#(X) 16] mark#(cons(X, Y)) =#> mark#(X) 17] mark#(s(X)) =#> mark#(X) Rules R_0: a!6220!6220app(nil, X) => mark(X) a!6220!6220app(cons(X, Y), Z) => cons(mark(X), app(Y, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220zWadr(nil, X) => nil a!6220!6220zWadr(X, nil) => nil a!6220!6220zWadr(cons(X, Y), cons(Z, U)) => cons(a!6220!6220app(mark(Z), cons(mark(X), nil)), zWadr(Y, U)) a!6220!6220prefix(X) => cons(nil, zWadr(X, prefix(X))) mark(app(X, Y)) => a!6220!6220app(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(zWadr(X, Y)) => a!6220!6220zWadr(mark(X), mark(Y)) mark(prefix(X)) => a!6220!6220prefix(mark(X)) mark(nil) => nil mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) a!6220!6220app(X, Y) => app(X, Y) a!6220!6220from(X) => from(X) a!6220!6220zWadr(X, Y) => zWadr(X, Y) a!6220!6220prefix(X) => prefix(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 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 1 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 2 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 3 : 0, 1 * 4 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17
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