TRS Stand 20472 pair #381711921

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	n030.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.9145128727 seconds
cpu usage	4.37474222
max memory	2.61541888E8

stage attributes

key	value
output-size	12124
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 47 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MRRProof [EQUIVALENT, 0 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

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