details

property	value
status	complete
benchmark	Applicative_AG01_innermost__#4.24.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.570835 seconds
cpu usage	0.482093
user time	0.416709
system time	0.065384
max virtual memory	127348.0
max residence set size	12020.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [b * c] --> c false : [] --> a filter : [b -> a * c] --> c filter2 : [a * b -> a * b * c] --> c int : [b * b] --> c intlist : [c] --> c map : [b -> b * c] --> c nil : [] --> c s : [b] --> b true : [] --> a Rules: intlist(nil) => nil intlist(cons(x, y)) => cons(s(x), intlist(y)) int(0, 0) => cons(0, nil) int(0, s(x)) => cons(0, int(s(0), s(x))) int(s(x), 0) => nil int(s(x), s(y)) => intlist(int(x, y)) 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: intlist(nil) => nil intlist(cons(X, Y)) => cons(s(X), intlist(Y)) int(0, 0) => cons(0, nil) int(0, s(X)) => cons(0, int(s(0), s(X))) int(s(X), 0) => nil int(s(X), s(Y)) => intlist(int(X, Y)) Moreover, the system is orthogonal. 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 terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || Input TRS: || 1: intlist(nil()) -> nil() || 2: intlist(cons(PeRCenTX,PeRCenTY)) -> cons(s(PeRCenTX),intlist(PeRCenTY)) || 3: int(0(),0()) -> cons(0(),nil()) || 4: int(0(),s(PeRCenTX)) -> cons(0(),int(s(0()),s(PeRCenTX))) || 5: int(s(PeRCenTX),0()) -> nil() || 6: int(s(PeRCenTX),s(PeRCenTY)) -> intlist(int(PeRCenTX,PeRCenTY)) || Number of strict rules: 6 || Direct POLO(bPol) ... removes: 3 5 || s w: x1 || 0 w: 0 || nil w: 1 || intlist w: x1 || int w: x1 + x2 + 2 || cons w: x1 + x2 || Number of strict rules: 4 || Direct POLO(bPol) ... removes: 1 || s w: 2 * x1 || 0 w: 0 || nil w: 1 || intlist w: 2 * x1 || int w: x1 + x2 || cons w: x1 + x2 || Number of strict rules: 3 || Direct POLO(bPol) ... failed. || Uncurrying int || 2: intlist(cons(PeRCenTX,PeRCenTY)) -> cons(s(PeRCenTX),intlist(PeRCenTY)) || 4: int^1_0(s(PeRCenTX)) -> cons(0(),int^1_s(0(),s(PeRCenTX))) || 6: int^1_s(PeRCenTX,s(PeRCenTY)) -> intlist(int(PeRCenTX,PeRCenTY)) || 7: int(0(),_1) ->= int^1_0(_1) || 8: int(s(_1),_2) ->= int^1_s(_1,_2) || Number of strict rules: 3 || Direct POLO(bPol) ... removes: 8 || int^1_0 w: 2 * x1 + 1 || s w: 2 * x1 + 1 || 0 w: 0 || nil w: 0 || intlist w: 2 * x1 || int w: x1 + 2 * x2 + 1 || cons w: x1 + x2 + 1 || int^1_s w: 2 * x1 + 2 * x2 || Number of strict rules: 3 || Direct POLO(bPol) ... failed. || Dependency Pairs: || #1: #intlist(cons(PeRCenTX,PeRCenTY)) -> #intlist(PeRCenTY) || #2: #int^1_s(PeRCenTX,s(PeRCenTY)) -> #intlist(int(PeRCenTX,PeRCenTY)) || #3: #int^1_s(PeRCenTX,s(PeRCenTY)) -> #int(PeRCenTX,PeRCenTY) || #4: #int(0(),_1) ->? #int^1_0(_1) || #5: #int^1_0(s(PeRCenTX)) -> #int^1_s(0(),s(PeRCenTX)) || Number of SCCs: 2, DPs: 4 || SCC { #1 } || POLO(Sum)... succeeded. || int^1_0 w: 0 popout

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

job log

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actions

all output return to Higher Order Rewriting Union Beta