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TRS Stand 20472 pair #381709747
details
property
value
status
complete
benchmark
recursion-10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
TCT_12
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
1.20952701569 seconds
cpu usage
1.205971449
max memory
3.5192832E7
stage attributes
key
value
output-size
37473
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. a : [] --> o b : [o * o] --> o f!62200 : [o] --> o f!62201 : [o] --> o f!622010 : [o] --> o f!62202 : [o] --> o f!62203 : [o] --> o f!62204 : [o] --> o f!62205 : [o] --> o f!62206 : [o] --> o f!62207 : [o] --> o f!62208 : [o] --> o f!62209 : [o] --> o g!62201 : [o * o] --> o g!622010 : [o * o] --> o g!62202 : [o * o] --> o g!62203 : [o * o] --> o g!62204 : [o * o] --> o g!62205 : [o * o] --> o g!62206 : [o * o] --> o g!62207 : [o * o] --> o g!62208 : [o * o] --> o g!62209 : [o * o] --> o s : [o] --> o f!62200(X) => a f!62201(X) => g!62201(X, X) g!62201(s(X), Y) => b(f!62200(Y), g!62201(X, Y)) f!62202(X) => g!62202(X, X) g!62202(s(X), Y) => b(f!62201(Y), g!62202(X, Y)) f!62203(X) => g!62203(X, X) g!62203(s(X), Y) => b(f!62202(Y), g!62203(X, Y)) f!62204(X) => g!62204(X, X) g!62204(s(X), Y) => b(f!62203(Y), g!62204(X, Y)) f!62205(X) => g!62205(X, X) g!62205(s(X), Y) => b(f!62204(Y), g!62205(X, Y)) f!62206(X) => g!62206(X, X) g!62206(s(X), Y) => b(f!62205(Y), g!62206(X, Y)) f!62207(X) => g!62207(X, X) g!62207(s(X), Y) => b(f!62206(Y), g!62207(X, Y)) f!62208(X) => g!62208(X, X) g!62208(s(X), Y) => b(f!62207(Y), g!62208(X, Y)) f!62209(X) => g!62209(X, X) g!62209(s(X), Y) => b(f!62208(Y), g!62209(X, Y)) f!622010(X) => g!622010(X, X) g!622010(s(X), Y) => b(f!62209(Y), g!622010(X, Y)) 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: a : [] --> lh b : [lh * lh] --> lh f!62200 : [dh] --> lh f!62201 : [dh] --> lh f!622010 : [dh] --> lh f!62202 : [dh] --> lh f!62203 : [dh] --> lh f!62204 : [dh] --> lh f!62205 : [dh] --> lh f!62206 : [dh] --> lh f!62207 : [dh] --> lh f!62208 : [dh] --> lh f!62209 : [dh] --> lh g!62201 : [dh * dh] --> lh g!622010 : [dh * dh] --> lh g!62202 : [dh * dh] --> lh g!62203 : [dh * dh] --> lh g!62204 : [dh * dh] --> lh g!62205 : [dh * dh] --> lh g!62206 : [dh * dh] --> lh g!62207 : [dh * dh] --> lh g!62208 : [dh * dh] --> lh g!62209 : [dh * dh] --> lh s : [dh] --> dh 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!62200(X) >? a f!62201(X) >? g!62201(X, X) g!62201(s(X), Y) >? b(f!62200(Y), g!62201(X, Y)) f!62202(X) >? g!62202(X, X) g!62202(s(X), Y) >? b(f!62201(Y), g!62202(X, Y)) f!62203(X) >? g!62203(X, X) g!62203(s(X), Y) >? b(f!62202(Y), g!62203(X, Y)) f!62204(X) >? g!62204(X, X) g!62204(s(X), Y) >? b(f!62203(Y), g!62204(X, Y)) f!62205(X) >? g!62205(X, X) g!62205(s(X), Y) >? b(f!62204(Y), g!62205(X, Y))
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