TRS Innermost pair #487524573

details

property	value
status	complete
benchmark	Ex2_Luc02a_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n092.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.67248415947 seconds
cpu usage	3.882863488
max memory	2.35102208E8

stage attributes

key	value
output-size	16722
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 [SOUND, 0 ms]
(2) CSR
(3) CSRInnermostProof [EQUIVALENT, 0 ms]
(4) CSR
(5) CSDependencyPairsProof [EQUIVALENT, 12 ms]
(6) QCSDP
(7) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QCSDP
        (10) QCSDPSubtermProof [EQUIVALENT, 0 ms]
        (11) QCSDP
        (12) PIsEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QCSDP
        (15) QCSDPSubtermProof [EQUIVALENT, 0 ms]
        (16) QCSDP
        (17) PIsEmptyProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QCSDP
        (20) QCSDPSubtermProof [EQUIVALENT, 0 ms]
        (21) QCSDP
        (22) PIsEmptyProof [EQUIVALENT, 0 ms]
        (23) YES


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

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

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   active(terms(X)) -> terms(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(recip(X)) -> recip(active(X))
   active(sqr(X)) -> sqr(active(X))
   active(s(X)) -> s(active(X))
   active(add(X1, X2)) -> add(active(X1), X2)
   active(add(X1, X2)) -> add(X1, active(X2))
   active(dbl(X)) -> dbl(active(X))
   active(first(X1, X2)) -> first(active(X1), X2)
   active(first(X1, X2)) -> first(X1, active(X2))
   terms(mark(X)) -> mark(terms(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   recip(mark(X)) -> mark(recip(X))
   sqr(mark(X)) -> mark(sqr(X))
   s(mark(X)) -> mark(s(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   dbl(mark(X)) -> mark(dbl(X))
   first(mark(X1), X2) -> mark(first(X1, X2))
   first(X1, mark(X2)) -> mark(first(X1, X2))
   proper(terms(X)) -> terms(proper(X))
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(recip(X)) -> recip(proper(X))
   proper(sqr(X)) -> sqr(proper(X))
   proper(s(X)) -> s(proper(X))
   proper(0) -> ok(0)
   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
   proper(dbl(X)) -> dbl(proper(X))
   proper(first(X1, X2)) -> first(proper(X1), proper(X2))
   proper(nil) -> ok(nil)
   terms(ok(X)) -> ok(terms(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   recip(ok(X)) -> ok(recip(X))
   sqr(ok(X)) -> ok(sqr(X))
   s(ok(X)) -> ok(s(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   dbl(ok(X)) -> ok(dbl(X))
   first(ok(X1), ok(X2)) -> ok(first(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(terms(x0))
   active(cons(x0, x1))

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