details

property	value
status	complete
benchmark	secret5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	2.9737 seconds
cpu usage	10.3683
user time	8.65884
system time	1.70944
max virtual memory	5989384.0
max residence set size	101100.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: t(N) -> cs(r(q(N)),nt(ns(N))) q(0()) -> 0() q(s(X)) -> s(p(q(X),d(X))) d(0()) -> 0() d(s(X)) -> s(s(d(X))) p(0(),X) -> X p(X,0()) -> X p(s(X),s(Y)) -> s(s(p(X,Y))) f(0(),X) -> nil() f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z))) t(X) -> nt(X) s(X) -> ns(X) f(X1,X2) -> nf(X1,X2) a(nt(X)) -> t(a(X)) a(ns(X)) -> s(a(X)) a(nf(X1,X2)) -> f(a(X1),a(X2)) a(X) -> X Proof: WPO Processor: algebra: Max weight function: w0 = 0 w(nf) = 5 w(0) = 3 w(nil) = 2 w(f) = w(cs) = w(nt) = w(r) = w(q) = w(t) = 1 w(a) = w(p) = w(d) = w(s) = w(ns) = 0 status function: st(nf) = [0, 1] st(a) = [0] st(nil) = [] st(f) = st(p) = [0, 1] st(d) = st(s) = [0] st(0) = [] st(cs) = [0, 1] st(nt) = st(ns) = st(r) = st(q) = st(t) = [0] subterm penalty function: sp(nf, 1) = sp(f, 1) = 5 sp(nt, 0) = sp(q, 0) = sp(t, 0) = 1 sp(nf, 0) = sp(a, 0) = sp(f, 0) = sp(p, 1) = sp(p, 0) = sp(d, 0) = sp( s, 0) = sp(cs, 1) = sp(cs, 0) = sp(ns, 0) = sp(r, 0) = 0 precedence: a > t > q > p > d > s > f > nf ~ nil ~ 0 ~ cs ~ nt ~ ns ~ r problem: Qed popout

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

job log

popout

actions

all output return to TRS Standard