details

property	value
status	complete
benchmark	jones6.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n111.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0906610488892 seconds
cpu usage	0.036274267
max memory	1527808.0

stage attributes

key	value
output-size	3379
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. cons : [o * o] --> o empty : [] --> o f : [o * o] --> o g : [o * o] --> o f(X, empty) => g(X, empty) f(X, cons(Y, Z)) => f(cons(Y, X), Z) g(empty, X) => X g(cons(X, Y), Z) => g(Y, cons(X, Z)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: cons : [s * sa] --> sa empty : [] --> sa f : [sa * sa] --> sa g : [sa * sa] --> sa 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(X, empty) >? g(X, empty) f(X, cons(Y, Z)) >? f(cons(Y, X), Z) g(empty, X) >? X g(cons(X, Y), Z) >? g(Y, cons(X, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.y0 + y1 empty = 3 f = \y0y1.3 + 2y0 + 2y1 g = \y0y1.2y0 + 2y1 Using this interpretation, the requirements translate to: [[f(_x0, empty)]] = 9 + 2x0 > 6 + 2x0 = [[g(_x0, empty)]] [[f(_x0, cons(_x1, _x2))]] = 3 + 2x0 + 2x1 + 2x2 >= 3 + 2x0 + 2x1 + 2x2 = [[f(cons(_x1, _x0), _x2)]] [[g(empty, _x0)]] = 6 + 2x0 > x0 = [[_x0]] [[g(cons(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[g(_x1, cons(_x0, _x2))]] We can thus remove the following rules: f(X, empty) => g(X, empty) g(empty, X) => X 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(X, cons(Y, Z)) >? f(cons(Y, X), Z) g(cons(X, Y), Z) >? g(Y, cons(X, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + y1 + 3y0 f = \y0y1.y0 + 2y1 g = \y0y1.2y1 + 3y0 Using this interpretation, the requirements translate to: [[f(_x0, cons(_x1, _x2))]] = 6 + x0 + 2x2 + 6x1 > 3 + x0 + 2x2 + 3x1 = [[f(cons(_x1, _x0), _x2)]] [[g(cons(_x0, _x1), _x2)]] = 9 + 2x2 + 3x1 + 9x0 > 6 + 2x2 + 3x1 + 6x0 = [[g(_x1, cons(_x0, _x2))]] We can thus remove the following rules: f(X, cons(Y, Z)) => f(cons(Y, X), Z) g(cons(X, Y), Z) => g(Y, cons(X, Z)) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [FuhGieParSchSwi11] C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski. Proving Termination by Dependency Pairs and Inductive Theorem Proving. In volume 47(2) of Journal of Automated Reasoning. 133--160, 2011. [Gra95] B. Gramlich. Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems. In volume 24(1-2) of Fundamentae Informaticae. 3--23, 1995. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. popout

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