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TRS Stand 20472 pair #381712380
details
property
value
status
complete
benchmark
#3.12.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0887889862061 seconds
cpu usage
0.078296909
max memory
2211840.0
stage attributes
key
value
output-size
6517
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. add : [o * o] --> o app : [o * o] --> o nil : [] --> o reverse : [o] --> o shuffle : [o] --> o app(nil, X) => X app(add(X, Y), Z) => add(X, app(Y, Z)) reverse(nil) => nil reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) shuffle(nil) => nil shuffle(add(X, Y)) => add(X, shuffle(reverse(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: add : [q * fb] --> fb app : [fb * fb] --> fb nil : [] --> fb reverse : [fb] --> fb shuffle : [fb] --> fb We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(nil, X) >? X app(add(X, Y), Z) >? add(X, app(Y, Z)) reverse(nil) >? nil reverse(add(X, Y)) >? app(reverse(Y), add(X, nil)) shuffle(nil) >? nil shuffle(add(X, Y)) >? add(X, shuffle(reverse(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: add = \y0y1.y0 + y1 app = \y0y1.y0 + y1 nil = 0 reverse = \y0.y0 shuffle = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(add(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[add(_x0, app(_x1, _x2))]] [[reverse(nil)]] = 0 >= 0 = [[nil]] [[reverse(add(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[app(reverse(_x1), add(_x0, nil))]] [[shuffle(nil)]] = 2 > 0 = [[nil]] [[shuffle(add(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + x0 + 2x1 = [[add(_x0, shuffle(reverse(_x1)))]] We can thus remove the following rules: shuffle(nil) => nil We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(nil, X) >? X app(add(X, Y), Z) >? add(X, app(Y, Z)) reverse(nil) >? nil reverse(add(X, Y)) >? app(reverse(Y), add(X, nil)) shuffle(add(X, Y)) >? add(X, shuffle(reverse(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: add = \y0y1.1 + y1 + 2y0 app = \y0y1.y0 + y1 nil = 0 reverse = \y0.y0 shuffle = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(add(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[add(_x0, app(_x1, _x2))]] [[reverse(nil)]] = 0 >= 0 = [[nil]] [[reverse(add(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[app(reverse(_x1), add(_x0, nil))]] [[shuffle(add(_x0, _x1))]] = 4 + 2x1 + 4x0 > 3 + 2x0 + 2x1 = [[add(_x0, shuffle(reverse(_x1)))]] We can thus remove the following rules: shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) We use rule removal, following [Kop12, Theorem 2.23].
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