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TRS Stand 20472 pair #381710727
details
property
value
status
complete
benchmark
#3.53.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n024.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.295224189758 seconds
cpu usage
0.29306557
max memory
9965568.0
stage attributes
key
value
output-size
15426
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 add : [o * o] --> o app : [o * o] --> o concat : [o * o] --> o cons : [o * o] --> o false : [] --> o leaf : [] --> o less!6220leaves : [o * o] --> o minus : [o * o] --> o nil : [] --> o quot : [o * o] --> o reverse : [o] --> o s : [o] --> o shuffle : [o] --> o true : [] --> o minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 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))) concat(leaf, X) => X concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) less!6220leaves(X, leaf) => false less!6220leaves(leaf, cons(X, Y)) => true less!6220leaves(cons(X, Y), cons(Z, U)) => less!6220leaves(concat(X, Y), concat(Z, U)) 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: 0 : [] --> ze add : [lb * ze] --> ze app : [ze * ze] --> ze concat : [ze * ze] --> ze cons : [ze * ze] --> ze false : [] --> af leaf : [] --> ze less!6220leaves : [ze * ze] --> af minus : [ze * ze] --> ze nil : [] --> ze quot : [ze * ze] --> ze reverse : [ze] --> ze s : [ze] --> ze shuffle : [ze] --> ze true : [] --> af 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] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 2] quot#(s(X), s(Y)) =#> minus#(X, Y) 3] app#(add(X, Y), Z) =#> app#(Y, Z) 4] reverse#(add(X, Y)) =#> app#(reverse(Y), add(X, nil)) 5] reverse#(add(X, Y)) =#> reverse#(Y) 6] shuffle#(add(X, Y)) =#> shuffle#(reverse(Y)) 7] shuffle#(add(X, Y)) =#> reverse#(Y) 8] concat#(cons(X, Y), Z) =#> concat#(Y, Z) 9] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> less!6220leaves#(concat(X, Y), concat(Z, U)) 10] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> concat#(X, Y) 11] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> concat#(Z, U) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 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))) concat(leaf, X) => X concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) less!6220leaves(X, leaf) => false less!6220leaves(leaf, cons(X, Y)) => true less!6220leaves(cons(X, Y), cons(Z, U)) => less!6220leaves(concat(X, Y), concat(Z, U)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.
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