TRS Standard pair #487074068

details

property	value
status	complete
benchmark	filliatre3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.1202 seconds
cpu usage	9.72966
user time	9.30182
system time	0.427836
max virtual memory	1.8475476E7
max residence set size	283756.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 5 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 46 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MRRProof [EQUIVALENT, 2 ms]
        (15) QDP
        (16) PisEmptyProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   g(A) -> A
   g(B) -> A
   g(B) -> B
   g(C) -> A
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))
   f'(triple(a, b, c), C) -> triple(a, b, s(c))
   f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
   f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
   f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
   fold(t, x, 0) -> t
   fold(t, x, s(n)) -> f(fold(t, x, n), x)

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FOLDB(t, s(n)) -> F(foldB(t, n), B)
   FOLDB(t, s(n)) -> FOLDB(t, n)
   FOLDC(t, s(n)) -> F(foldC(t, n), C)
   FOLDC(t, s(n)) -> FOLDC(t, n)
   F(t, x) -> F'(t, g(x))
   F(t, x) -> G(x)
   F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
   F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
   F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
   F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
   FOLD(t, x, s(n)) -> F(fold(t, x, n), x)
   FOLD(t, x, s(n)) -> FOLD(t, x, n)

The TRS R consists of the following rules:

   g(A) -> A
   g(B) -> A
   g(B) -> B
   g(C) -> A
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))
   f'(triple(a, b, c), C) -> triple(a, b, s(c))
   f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
   f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
   f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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