details

property	value
status	complete
benchmark	jones1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0298240184784 seconds
cpu usage	0.026109537
max memory	1417216.0

stage attributes

key	value
output-size	2893
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 r1 : [o * o] --> o rev : [o] --> o rev(X) => r1(X, empty) r1(empty, X) => X r1(cons(X, Y), Z) => r1(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 : [v * fa] --> fa empty : [] --> fa r1 : [fa * fa] --> fa rev : [fa] --> fa We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): rev(X) >? r1(X, empty) r1(empty, X) >? X r1(cons(X, Y), Z) >? r1(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 empty = 0 r1 = \y0y1.2y1 + 3y0 rev = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[rev(_x0)]] = 3 + 3x0 > 3x0 = [[r1(_x0, empty)]] [[r1(empty, _x0)]] = 2x0 >= x0 = [[_x0]] [[r1(cons(_x0, _x1), _x2)]] = 9 + 2x2 + 3x1 + 9x0 > 6 + 2x2 + 3x1 + 6x0 = [[r1(_x1, cons(_x0, _x2))]] We can thus remove the following rules: rev(X) => r1(X, empty) r1(cons(X, Y), Z) => r1(Y, cons(X, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): r1(empty, X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: empty = 3 r1 = \y0y1.3 + y0 + y1 Using this interpretation, the requirements translate to: [[r1(empty, _x0)]] = 6 + x0 > x0 = [[_x0]] We can thus remove the following rules: r1(empty, X) => X 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

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

job log

popout

actions

all output return to TRS Stand 20472