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TRS Stand 20472 pair #381711079
details
property
value
status
complete
benchmark
hydra.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n028.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0590291023254 seconds
cpu usage
0.054067992
max memory
3375104.0
stage attributes
key
value
output-size
2909
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. 0 : [] --> o cons : [o * o] --> o copy : [o * o * o] --> o f : [o] --> o n : [] --> o nil : [] --> o s : [o] --> o f(cons(nil, X)) => X f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) copy(0, X, Y) => f(Y) copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] f#(cons(f(cons(nil, X)), Y)) =#> copy#(n, X, Y) 1] copy#(0, X, Y) =#> f#(Y) 2] copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z)) 3] copy#(s(X), Y, Z) =#> f#(Y) Rules R_0: f(cons(nil, X)) => X f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) copy(0, X, Y) => f(Y) copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 0 * 2 : 1, 2, 3 * 3 : 0 This graph has the following strongly connected components: P_1: copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z)) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(copy#) = 1 Thus, we can orient the dependency pairs as follows: nu(copy#(s(X), Y, Z)) = s(X) |> X = nu(copy#(X, Y, cons(f(Y), Z))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.
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