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	n072.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	0.588119029999 seconds
cpu usage	1.141402766
max memory	5.7106432E7

stage attributes

key	value
output-size	6649
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) QTRS Reverse [EQUIVALENT] popout

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job log

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actions

all output return to HRS union beta 16688