details

property	value
status	complete
benchmark	big.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n045.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.36514496803 seconds
cpu usage	1.35383922
max memory	3.3906688E7

stage attributes

key	value
output-size	37386
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. !940 : [] --> o !minus : [o * o] --> o !plus : [o * o] --> o !times : [o * o] --> o 0 : [o] --> o 1 : [o] --> o app : [o * o] --> o cons : [o * o] --> o eq : [o * o] --> o false : [] --> o ge : [o * o] --> o if : [o * o * o] --> o ifinter : [o * o * o * o] --> o inter : [o * o] --> o log : [o] --> o log!450 : [o] --> o mem : [o * o] --> o nil : [] --> o not : [o] --> o prod : [o] --> o sum : [o] --> o true : [] --> o 0(!940) => !940 !plus(X, !940) => X !plus(!940, X) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !minus(!940, X) => !940 !minus(X, !940) => X !minus(0(X), 0(Y)) => 0(!minus(X, Y)) !minus(0(X), 1(Y)) => 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) => 1(!minus(X, Y)) !minus(1(X), 1(Y)) => 0(!minus(X, Y)) not(true) => false not(false) => true if(true, X, Y) => X if(false, X, Y) => Y eq(!940, !940) => true eq(!940, 1(X)) => false eq(1(X), !940) => false eq(!940, 0(X)) => eq(!940, X) eq(0(X), !940) => eq(X, !940) eq(1(X), 1(Y)) => eq(X, Y) eq(0(X), 1(Y)) => false eq(1(X), 0(Y)) => false eq(0(X), 0(Y)) => eq(X, Y) ge(0(X), 0(Y)) => ge(X, Y) ge(0(X), 1(Y)) => not(ge(Y, X)) ge(1(X), 0(Y)) => ge(X, Y) ge(1(X), 1(Y)) => ge(X, Y) ge(X, !940) => true ge(!940, 0(X)) => ge(!940, X) ge(!940, 1(X)) => false log(X) => !minus(log!450(X), 1(!940)) log!450(!940) => !940 log!450(1(X)) => !plus(log!450(X), 1(!940)) log!450(0(X)) => if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) !times(!940, X) => !940 !times(0(X), Y) => 0(!times(X, Y)) !times(1(X), Y) => !plus(0(!times(X, Y)), Y) !times(!times(X, Y), Z) => !times(X, !times(Y, Z)) !times(X, !plus(Y, Z)) => !plus(!times(X, Y), !times(X, Z)) app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(nil) => 0(!940) sum(cons(X, Y)) => !plus(X, sum(Y)) sum(app(X, Y)) => !plus(sum(X), sum(Y)) prod(nil) => 1(!940) prod(cons(X, Y)) => !times(X, prod(Y)) prod(app(X, Y)) => !times(prod(X), prod(Y)) mem(X, nil) => false mem(X, cons(Y, Z)) => if(eq(X, Y), true, mem(X, Z)) 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): popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Stand 20472