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TRS Stand 20472 pair #381716625
details
property
value
status
complete
benchmark
Ex1_2_Luc02c_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n048.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.240799188614 seconds
cpu usage
0.20882438
max memory
4837376.0
stage attributes
key
value
output-size
14094
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. 2nd : [o] --> o active : [o] --> o cons : [o * o] --> o from : [o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o active(2nd(cons(X, cons(Y, Z)))) => mark(Y) active(from(X)) => mark(cons(X, from(s(X)))) active(2nd(X)) => 2nd(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(from(X)) => from(active(X)) active(s(X)) => s(active(X)) 2nd(mark(X)) => mark(2nd(X)) cons(mark(X), Y) => mark(cons(X, Y)) from(mark(X)) => mark(from(X)) s(mark(X)) => mark(s(X)) proper(2nd(X)) => 2nd(proper(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(from(X)) => from(proper(X)) proper(s(X)) => s(proper(X)) 2nd(ok(X)) => ok(2nd(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) from(ok(X)) => ok(from(X)) s(ok(X)) => ok(s(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(X)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): active(2nd(cons(X, cons(Y, Z)))) >? mark(Y) active(from(X)) >? mark(cons(X, from(s(X)))) active(2nd(X)) >? 2nd(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) 2nd(ok(X)) >? ok(2nd(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) from(ok(X)) >? ok(from(X)) s(ok(X)) >? ok(s(X)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.2y0 cons = \y0y1.y0 + y1 from = \y0.y0 mark = \y0.y0 ok = \y0.1 + 2y0 proper = \y0.y0 s = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(2nd(cons(_x0, cons(_x1, _x2))))]] = 2x0 + 2x1 + 2x2 >= x1 = [[mark(_x1)]] [[active(from(_x0))]] = 2x0 >= 2x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(2nd(_x0))]] = 2x0 >= 2x0 = [[2nd(active(_x0))]] [[active(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 2x0 >= 2x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[2nd(mark(_x0))]] = x0 >= x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = x0 >= x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[proper(2nd(_x0))]] = x0 >= x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = x0 >= x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[2nd(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(2nd(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[from(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(from(_x0))]]
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