TRS Innermost pair #487093011

details

property	value
status	complete
benchmark	Ex6_9_Luc02c_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.81822 seconds
cpu usage	4.05237
user time	3.8646
system time	0.187767
max virtual memory	1.8408912E7
max residence set size	272660.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 69 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 2 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
   active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(2nd(X)) -> 2nd(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(from(X)) -> from(active(X))
   active(s(X)) -> s(active(X))
   active(cons1(X1, X2)) -> cons1(active(X1), X2)
   active(cons1(X1, X2)) -> cons1(X1, active(X2))
   2nd(mark(X)) -> mark(2nd(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   from(mark(X)) -> mark(from(X))
   s(mark(X)) -> mark(s(X))
   cons1(mark(X1), X2) -> mark(cons1(X1, X2))
   cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
   proper(2nd(X)) -> 2nd(proper(X))
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(from(X)) -> from(proper(X))
   proper(s(X)) -> s(proper(X))
   proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
   2nd(ok(X)) -> ok(2nd(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   from(ok(X)) -> ok(from(X))
   s(ok(X)) -> ok(s(X))
   cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(from(x0))
   active(2nd(x0))
   active(cons(x0, x1))
   active(s(x0))
   active(cons1(x0, x1))
   2nd(mark(x0))
   cons(mark(x0), x1)
   from(mark(x0))
   s(mark(x0))
   cons1(mark(x0), x1)
   cons1(x0, mark(x1))
   proper(2nd(x0))
   proper(cons(x0, x1))
   proper(from(x0))
   proper(s(x0))
   proper(cons1(x0, x1))
   2nd(ok(x0))
   cons(ok(x0), ok(x1))
   from(ok(x0))
   s(ok(x0))
   cons1(ok(x0), ok(x1))
   top(mark(x0))
   top(ok(x0))


----------------------------------------

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(2nd(x_1)) = 1 + 2*x_1
   POL(active(x_1)) = 1 + 2*x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(cons1(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(from(x_1)) = 2 + 2*x_1
   POL(mark(x_1)) = 1 + x_1
   POL(ok(x_1)) = 2 + 2*x_1
   POL(proper(x_1)) = x_1
   POL(s(x_1)) = x_1
   POL(top(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
   active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(cons(X1, X2)) -> cons(active(X1), X2)

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