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TRS Stand 20472 pair #381714225
details
property
value
status
complete
benchmark
2.39.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n108.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0739870071411 seconds
cpu usage
0.056950167
max memory
1814528.0
stage attributes
key
value
output-size
6043
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. !dot : [o * o] --> o !plus!plus : [o * o] --> o car : [o] --> o cdr : [o] --> o false : [] --> o nil : [] --> o null : [o] --> o rev : [o] --> o true : [] --> o rev(nil) => nil rev(!dot(X, Y)) => !plus!plus(rev(Y), !dot(X, nil)) car(!dot(X, Y)) => X cdr(!dot(X, Y)) => Y null(nil) => true null(!dot(X, Y)) => false !plus!plus(nil, X) => X !plus!plus(!dot(X, Y), Z) => !dot(X, !plus!plus(Y, 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: !dot : [q * tb] --> tb !plus!plus : [tb * tb] --> tb car : [tb] --> q cdr : [tb] --> tb false : [] --> cb nil : [] --> tb null : [tb] --> cb rev : [tb] --> tb true : [] --> cb 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(nil) >? nil rev(!dot(X, Y)) >? !plus!plus(rev(Y), !dot(X, nil)) car(!dot(X, Y)) >? X cdr(!dot(X, Y)) >? Y null(nil) >? true null(!dot(X, Y)) >? false !plus!plus(nil, X) >? X !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y0 + y1 !plus!plus = \y0y1.y0 + 2y1 car = \y0.3 + 2y0 cdr = \y0.3 + 2y0 false = 0 nil = 0 null = \y0.3 + 3y0 rev = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[rev(nil)]] = 0 >= 0 = [[nil]] [[rev(!dot(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(rev(_x1), !dot(_x0, nil))]] [[car(!dot(_x0, _x1))]] = 3 + 2x0 + 2x1 > x0 = [[_x0]] [[cdr(!dot(_x0, _x1))]] = 3 + 2x0 + 2x1 > x1 = [[_x1]] [[null(nil)]] = 3 > 0 = [[true]] [[null(!dot(_x0, _x1))]] = 3 + 3x0 + 3x1 > 0 = [[false]] [[!plus!plus(nil, _x0)]] = 2x0 >= x0 = [[_x0]] [[!plus!plus(!dot(_x0, _x1), _x2)]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[!dot(_x0, !plus!plus(_x1, _x2))]] We can thus remove the following rules: car(!dot(X, Y)) => X cdr(!dot(X, Y)) => Y null(nil) => true null(!dot(X, Y)) => false 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(nil) >? nil rev(!dot(X, Y)) >? !plus!plus(rev(Y), !dot(X, nil)) !plus!plus(nil, X) >? X !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements:
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