details

property	value
status	complete
benchmark	Applicative_first_order_05__#3.25.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)	0.245325 seconds
cpu usage	0.236639
user time	0.197446
system time	0.039193
max virtual memory	113188.0
max residence set size	5068.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: cons : [c * d] --> d f : [a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d g : [a] --> a h : [a] --> a map : [c -> c * d] --> d nil : [] --> d true : [] --> b Rules: f(g(x)) => g(f(f(x))) f(h(x)) => h(g(x)) map(i, nil) => nil map(i, cons(x, y)) => cons(i x, map(i, y)) filter(i, nil) => nil filter(i, cons(x, y)) => filter2(i x, i, x, y) filter2(true, i, x, y) => cons(x, filter(i, y)) filter2(false, i, x, y) => filter(i, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(g(X)) >? g(f(f(X))) f(h(X)) >? h(g(X)) 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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.1 + y0 + y1 f = \y0.y0 false = 3 filter = \G0y1.2 + 2y1 + G0(0) + 2y1G0(y1) filter2 = \y0G1y2y3.1 + y0 + y2 + 2y3 + G1(0) + 2y3G1(y3) g = \y0.y0 h = \y0.y0 map = \G0y1.2 + 3y1 + 2y1G0(y1) + 2G0(y1) nil = 0 true = 3 Using this interpretation, the requirements translate to: [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] [[map(_F0, nil)]] = 2 + 2F0(0) > 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] We can thus remove the following rules: 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) We observe that the rules contain a first-order subset: f(g(X)) => g(f(f(X))) f(h(X)) => h(g(X)) 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: f(g(PeRCenTX)) -> g(f(f(PeRCenTX))) || 2: f(h(PeRCenTX)) -> h(g(PeRCenTX)) || Number of strict rules: 2 || Direct POLO(bPol) ... failed. || Uncurrying ... failed. || Dependency Pairs: || #1: #f(g(PeRCenTX)) -> #f(f(PeRCenTX)) || #2: #f(g(PeRCenTX)) -> #f(PeRCenTX) || Number of SCCs: 1, DPs: 2 || SCC { #1 #2 } || POLO(Sum)... succeeded. || h w: 1 || f w: x1 popout

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job log

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actions

all output return to Higher Order Rewriting Union Beta