details

property	value
status	complete
benchmark	Applicative_AG01_innermost__#4.17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n039.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	sol 37957
configuration	hrs
runtime (wallclock)	0.365966081619 seconds
cpu usage	0.408769354
max memory	6.7371008E7

stage attributes

key	value
output-size	2259
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_hrs /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO We split firstr-order part and higher-order part, and do modular checking by a general modularity. ******** FO SN check ******** Check non-SN using NTI (Non-Termination Inference by Payet) ******** Computation rules ******** (1) f(g(X),s(0),Y) => f(g(s(0)),Y,g(X)) (2) g(s(U)) => s(g(U)) (3) g(0) => 0 forward process killed -- 22 rule(s) generated forward+backward process killed -- 26 rule(s) generated NO let R be the TRS under consideration f(g(_1),s(0),_2) -> f(g(s(0)),_2,g(_1)) is in elim_R(R) let l0 be the left-hand side of this rule p0 = 1.0 is a position in l0 we have l0|p0 = 0 g(0) -> 0 is in R let r'0 be the right-hand side of this rule theta0 = {} is a mgu of l0|p0 and r'0 ==> f(g(_1),s(g(0)),_2) -> f(g(s(0)),_2,g(_1)) is in EU_R^1 let l1 be the left-hand side of this rule p1 = 1 is a position in l1 we have l1|p1 = s(g(0)) g(s(_3)) -> s(g(_3)) is in R let r'1 be the right-hand side of this rule theta1 = {_3/0} is a mgu of l1|p1 and r'1 ==> f(g(_1),g(s(0)),_2) -> f(g(s(0)),_2,g(_1)) is in EU_R^2 let l be the left-hand side and r be the right-hand side of this rule let p = epsilon let theta = {_1/s(0), _2/g(s(0))} let theta' = {} we have r|p = f(g(s(0)),_2,g(_1)) and theta'(theta(l)) = theta(r|p) so, theta(l) = f(g(s(0)),g(s(0)),g(s(0))) is non-terminating w.r.t. R Termination disproved by the backward process proof stopped at iteration i=2, depth k=3 15 rule(s) generated >>NO ******** Signature ******** 0 : b cons : (d,e) -> e f : (b,b,b) -> a false : c filter : ((d -> c),e) -> e filter2 : (c,(d -> c),d,e) -> e g : b -> b map : ((d -> d),e) -> e nil : e s : b -> b true : c ******** Computation Rules ******** (1) f(g(X),s(0),Y) => f(g(s(0)),Y,g(X)) (2) g(s(U)) => s(g(U)) (3) g(0) => 0 (4) map(H,nil) => nil (5) map(I,cons(P,X1)) => cons(I[P],map(I,X1)) (6) filter(Z1,nil) => nil (7) filter(G1,cons(V1,W1)) => filter2(G1[V1],G1,V1,W1) (8) filter2(true,J1,X2,Y2) => cons(X2,filter(J1,Y2)) (9) filter2(false,G2,V2,W2) => filter(G2,W2) NO popout

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

job log

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actions

all output return to HRS 58631