TRS Standard pair #487069363

details

property	value
status	complete
benchmark	ExProp7_Luc06_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.96241 seconds
cpu usage	3.86404
user time	3.68966
system time	0.17438
max virtual memory	1.840914E7
max residence set size	234844.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSToCSRProof [EQUIVALENT, 0 ms]
(2) CSR
(3) CSRRRRProof [EQUIVALENT, 27 ms]
(4) CSR
(5) CSRRRRProof [EQUIVALENT, 4 ms]
(6) CSR
(7) CSRRRRProof [EQUIVALENT, 0 ms]
(8) CSR
(9) RisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

   active(f(0)) -> mark(cons(0, f(s(0))))
   active(f(s(0))) -> mark(f(p(s(0))))
   active(p(s(X))) -> mark(X)
   active(f(X)) -> f(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(s(X)) -> s(active(X))
   active(p(X)) -> p(active(X))
   f(mark(X)) -> mark(f(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   s(mark(X)) -> mark(s(X))
   p(mark(X)) -> mark(p(X))
   proper(f(X)) -> f(proper(X))
   proper(0) -> ok(0)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(s(X)) -> s(proper(X))
   proper(p(X)) -> p(proper(X))
   f(ok(X)) -> ok(f(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   s(ok(X)) -> ok(s(X))
   p(ok(X)) -> ok(p(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.

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

(1) QTRSToCSRProof (EQUIVALENT)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

   active(f(0)) -> mark(cons(0, f(s(0))))
   active(f(s(0))) -> mark(f(p(s(0))))
   active(p(s(X))) -> mark(X)
   active(f(X)) -> f(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(s(X)) -> s(active(X))
   active(p(X)) -> p(active(X))
   f(mark(X)) -> mark(f(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   s(mark(X)) -> mark(s(X))
   p(mark(X)) -> mark(p(X))
   proper(f(X)) -> f(proper(X))
   proper(0) -> ok(0)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(s(X)) -> s(proper(X))
   proper(p(X)) -> p(proper(X))
   f(ok(X)) -> ok(f(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   s(ok(X)) -> ok(s(X))
   p(ok(X)) -> ok(p(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
The replacement map contains the following entries:

f: {1}
0: empty set
cons: {1}
s: {1}
p: {1}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
----------------------------------------

(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   f(0) -> cons(0, f(s(0)))

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