details

property	value
status	complete
benchmark	intersect.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n113.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.23658490181 seconds
cpu usage	0.232978527
max memory	1.0432512E7

stage attributes

key	value
output-size	12060
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 app : [o * o] --> o cons : [o * o] --> o eq : [o * o] --> o false : [] --> o if : [o * o * o] --> o ifinter : [o * o * o * o] --> o ifmem : [o * o * o] --> o inter : [o * o] --> o mem : [o * o] --> o nil : [] --> o s : [o] --> o true : [] --> o if(true, X, Y) => X if(false, X, Y) => Y eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) app(app(X, Y), Z) => app(X, app(Y, Z)) mem(X, nil) => false mem(X, cons(Y, Z)) => ifmem(eq(X, Y), X, Z) ifmem(true, X, Y) => true ifmem(false, X, Y) => mem(X, Y) inter(X, nil) => nil inter(nil, X) => nil inter(app(X, Y), Z) => app(inter(X, Z), inter(Y, Z)) inter(X, app(Y, Z)) => app(inter(X, Y), inter(X, Z)) inter(cons(X, Y), Z) => ifinter(mem(X, Z), X, Y, Z) inter(X, cons(Y, Z)) => ifinter(mem(Y, X), Y, Z, X) ifinter(true, X, Y, Z) => cons(X, inter(Y, Z)) ifinter(false, X, Y, Z) => inter(Y, Z) 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] eq#(s(X), s(Y)) =#> eq#(X, Y) 1] app#(cons(X, Y), Z) =#> app#(Y, Z) 2] app#(app(X, Y), Z) =#> app#(X, app(Y, Z)) 3] app#(app(X, Y), Z) =#> app#(Y, Z) 4] mem#(X, cons(Y, Z)) =#> ifmem#(eq(X, Y), X, Z) 5] mem#(X, cons(Y, Z)) =#> eq#(X, Y) 6] ifmem#(false, X, Y) =#> mem#(X, Y) 7] inter#(app(X, Y), Z) =#> app#(inter(X, Z), inter(Y, Z)) 8] inter#(app(X, Y), Z) =#> inter#(X, Z) 9] inter#(app(X, Y), Z) =#> inter#(Y, Z) 10] inter#(X, app(Y, Z)) =#> app#(inter(X, Y), inter(X, Z)) 11] inter#(X, app(Y, Z)) =#> inter#(X, Y) 12] inter#(X, app(Y, Z)) =#> inter#(X, Z) 13] inter#(cons(X, Y), Z) =#> ifinter#(mem(X, Z), X, Y, Z) 14] inter#(cons(X, Y), Z) =#> mem#(X, Z) 15] inter#(X, cons(Y, Z)) =#> ifinter#(mem(Y, X), Y, Z, X) 16] inter#(X, cons(Y, Z)) =#> mem#(Y, X) 17] ifinter#(true, X, Y, Z) =#> inter#(Y, Z) 18] ifinter#(false, X, Y, Z) =#> inter#(Y, Z) Rules R_0: if(true, X, Y) => X if(false, X, Y) => Y eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) app(app(X, Y), Z) => app(X, app(Y, Z)) mem(X, nil) => false mem(X, cons(Y, Z)) => ifmem(eq(X, Y), X, Z) ifmem(true, X, Y) => true ifmem(false, X, Y) => mem(X, Y) inter(X, nil) => nil inter(nil, X) => nil inter(app(X, Y), Z) => app(inter(X, Z), inter(Y, Z)) inter(X, app(Y, Z)) => app(inter(X, Y), inter(X, Z)) inter(cons(X, Y), Z) => ifinter(mem(X, Z), X, Y, Z) inter(X, cons(Y, Z)) => ifinter(mem(Y, X), Y, Z, X) ifinter(true, X, Y, Z) => cons(X, inter(Y, Z)) ifinter(false, X, Y, Z) => inter(Y, 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). popout

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