details

property	value
status	complete
benchmark	Applicative_first_order_05__#3.57.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.902857 seconds
cpu usage	0.872741
user time	0.798072
system time	0.074669
max virtual memory	135728.0
max residence set size	20052.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> b app : [c * c] --> c cons : [b * c] --> c false : [] --> a filter : [b -> a * c] --> c filter2 : [a * b -> a * b * c] --> c map : [b -> b * c] --> c minus : [b * b] --> b nil : [] --> c plus : [b * b] --> b quot : [b * b] --> b s : [b] --> b sum : [c] --> c true : [] --> a Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) minus(minus(x, y), z) => minus(x, plus(y, z)) quot(0, s(x)) => 0 quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) app(nil, x) => x app(x, nil) => x app(cons(x, y), z) => cons(x, app(y, z)) sum(cons(x, nil)) => cons(x, nil) sum(cons(x, cons(y, z))) => sum(cons(plus(x, y), z)) sum(app(x, cons(y, cons(z, u)))) => sum(app(x, sum(cons(y, cons(z, u))))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) Moreover, the system is finitely branching. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed Ce-terminating: || Input TRS: || 1: minus(PeRCenTX,0()) -> PeRCenTX || 2: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(PeRCenTX,PeRCenTY) || 3: minus(minus(PeRCenTX,PeRCenTY),PeRCenTZ) -> minus(PeRCenTX,plus(PeRCenTY,PeRCenTZ)) || 4: quot(0(),s(PeRCenTX)) -> 0() || 5: quot(s(PeRCenTX),s(PeRCenTY)) -> s(quot(minus(PeRCenTX,PeRCenTY),s(PeRCenTY))) || 6: plus(0(),PeRCenTX) -> PeRCenTX || 7: plus(s(PeRCenTX),PeRCenTY) -> s(plus(PeRCenTX,PeRCenTY)) || 8: app(nil(),PeRCenTX) -> PeRCenTX || 9: app(PeRCenTX,nil()) -> PeRCenTX || 10: app(cons(PeRCenTX,PeRCenTY),PeRCenTZ) -> cons(PeRCenTX,app(PeRCenTY,PeRCenTZ)) || 11: sum(cons(PeRCenTX,nil())) -> cons(PeRCenTX,nil()) || 12: sum(cons(PeRCenTX,cons(PeRCenTY,PeRCenTZ))) -> sum(cons(plus(PeRCenTX,PeRCenTY),PeRCenTZ)) || 13: sum(app(PeRCenTX,cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU)))) -> sum(app(PeRCenTX,sum(cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU))))) || 14: TIlDePAIR(PeRCenTX,PeRCenTY) -> PeRCenTX || 15: TIlDePAIR(PeRCenTX,PeRCenTY) -> PeRCenTY || Number of strict rules: 15 || Direct POLO(bPol) ... failed. || Uncurrying ... failed. || Dependency Pairs: || #1: #minus(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY) || #2: #sum(app(PeRCenTX,cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU)))) -> #sum(app(PeRCenTX,sum(cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU))))) || #3: #sum(app(PeRCenTX,cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU)))) -> #app(PeRCenTX,sum(cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU)))) || #4: #sum(app(PeRCenTX,cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU)))) -> #sum(cons(PeRCenTY,cons(PeRCenTZ,PeRCenTU))) || #5: #sum(cons(PeRCenTX,cons(PeRCenTY,PeRCenTZ))) -> #sum(cons(plus(PeRCenTX,PeRCenTY),PeRCenTZ)) || #6: #sum(cons(PeRCenTX,cons(PeRCenTY,PeRCenTZ))) -> #plus(PeRCenTX,PeRCenTY) || #7: #plus(s(PeRCenTX),PeRCenTY) -> #plus(PeRCenTX,PeRCenTY) || #8: #app(cons(PeRCenTX,PeRCenTY),PeRCenTZ) -> #app(PeRCenTY,PeRCenTZ) || #9: #quot(s(PeRCenTX),s(PeRCenTY)) -> #quot(minus(PeRCenTX,PeRCenTY),s(PeRCenTY)) || #10: #quot(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY) || #11: #minus(minus(PeRCenTX,PeRCenTY),PeRCenTZ) -> #minus(PeRCenTX,plus(PeRCenTY,PeRCenTZ)) || #12: #minus(minus(PeRCenTX,PeRCenTY),PeRCenTZ) -> #plus(PeRCenTY,PeRCenTZ) || Number of SCCs: 6, DPs: 7 || SCC { #7 } || POLO(Sum)... succeeded. || TIlDePAIR w: 0 popout

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job log

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actions

all output return to Higher Order Rewriting Union Beta