details

property	value
status	complete
benchmark	filliatre3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n029.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.268606901169 seconds
cpu usage	0.263939401
max memory	1.1096064E7

stage attributes

key	value
output-size	11524
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. 0 : [] --> o A : [] --> o B : [] --> o C : [] --> o f : [o * o] --> o f!450 : [o * o] --> o f!450!450 : [o] --> o fold : [o * o * o] --> o foldB : [o * o] --> o foldC : [o * o] --> o g : [o] --> o s : [o] --> o triple : [o * o * o] --> o g(A) => A g(B) => A g(B) => B g(C) => A g(C) => B g(C) => C foldB(X, 0) => X foldB(X, s(Y)) => f(foldB(X, Y), B) foldC(X, 0) => X foldC(X, s(Y)) => f(foldC(X, Y), C) f(X, Y) => f!450(X, g(Y)) f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) fold(X, Y, 0) => X fold(X, Y, s(Z)) => f(fold(X, Y, Z), Y) 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] foldB#(X, s(Y)) =#> f#(foldB(X, Y), B) 1] foldB#(X, s(Y)) =#> foldB#(X, Y) 2] foldC#(X, s(Y)) =#> f#(foldC(X, Y), C) 3] foldC#(X, s(Y)) =#> foldC#(X, Y) 4] f#(X, Y) =#> f!450#(X, g(Y)) 5] f#(X, Y) =#> g#(Y) 6] f!450#(triple(X, Y, Z), B) =#> f#(triple(X, Y, Z), A) 7] f!450#(triple(X, Y, Z), A) =#> f!450!450#(foldB(triple(s(X), 0, Z), Y)) 8] f!450#(triple(X, Y, Z), A) =#> foldB#(triple(s(X), 0, Z), Y) 9] f!450!450#(triple(X, Y, Z)) =#> foldC#(triple(X, Y, 0), Z) 10] fold#(X, Y, s(Z)) =#> f#(fold(X, Y, Z), Y) 11] fold#(X, Y, s(Z)) =#> fold#(X, Y, Z) Rules R_0: g(A) => A g(B) => A g(B) => B g(C) => A g(C) => B g(C) => C foldB(X, 0) => X foldB(X, s(Y)) => f(foldB(X, Y), B) foldC(X, 0) => X foldC(X, s(Y)) => f(foldC(X, Y), C) f(X, Y) => f!450(X, g(Y)) f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) fold(X, Y, 0) => X fold(X, Y, s(Z)) => f(fold(X, Y, Z), 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 : 4, 5 * 1 : 0, 1 * 2 : 4, 5 * 3 : 2, 3 * 4 : 6, 7, 8 * 5 : * 6 : 4, 5 * 7 : 9 * 8 : 0, 1 * 9 : 2, 3 * 10 : 4, 5 * 11 : 10, 11 popout

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actions

all output return to TRS Stand 20472