TRS Standard pair #487071328

details

property	value
status	complete
benchmark	Ex4_7_77_Bor03_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.6744 seconds
cpu usage	3.51406
user time	3.35944
system time	0.154623
max virtual memory	1.8343604E7
max residence set size	233296.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 37 ms]
(4) CSR
(5) CSRRRRProof [EQUIVALENT, 0 ms]
(6) CSR
(7) RisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

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

   active(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(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(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(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:

zeros: empty set
cons: {1}
0: empty set
tail: {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:

   zeros -> cons(0, zeros)
   tail(cons(X, XS)) -> XS

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
tail: {1}

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

(3) CSRRRRProof (EQUIVALENT)
The following CSR is given: Context-sensitive rewrite system:
The TRS R consists of the following rules:

   zeros -> cons(0, zeros)
   tail(cons(X, XS)) -> XS

The replacement map contains the following entries:

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