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TRS Stand 20472 pair #381710746
details
property
value
status
complete
benchmark
#4.22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n050.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0535368919373 seconds
cpu usage
0.049914852
max memory
2834432.0
stage attributes
key
value
output-size
2565
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. 0 : [] --> o quot : [o * o * o] --> o s : [o] --> o quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) 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] quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) 1] quot#(X, 0, s(Y)) =#> quot#(X, s(Y), s(Y)) Rules R_0: quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) 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 apply the subterm criterion with the following projection function: nu(quot#) = 1 Thus, we can orient the dependency pairs as follows: nu(quot#(s(X), s(Y), Z)) = s(X) |> X = nu(quot#(X, Y, Z)) nu(quot#(X, 0, s(Y))) = X = X = nu(quot#(X, s(Y), s(Y))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_0, R_0, minimal, f) by (P_1, R_0, minimal, f), where P_1 contains: quot#(X, 0, s(Y)) =#> quot#(X, s(Y), s(Y)) 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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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