TRS Standard pair #487066893

details

property	value
status	complete
benchmark	worm_wrong.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n174.star.cs.uiowa.edu
space	Hydras

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	20.9343 seconds
cpu usage	59.5935
user time	57.1443
system time	2.44914
max virtual memory	3.6984768E7
max residence set size	4809280.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) TransformationProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 147 ms]
        (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 79 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 38 ms]
        (28) QDP
        (29) QDPOrderProof [EQUIVALENT, 2511 ms]
        (30) QDP
        (31) PisEmptyProof [EQUIVALENT, 0 ms]
        (32) YES


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

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

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

Q is empty.

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(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:

   .^1(.(x, y), z) -> .^1(x, .(y, z))
   .^1(.(x, y), z) -> .^1(y, z)
   A(f(x)) -> F(a(x))
   A(f(x)) -> A(x)
   A(.(x, y)) -> .^1(a(x), y)
   A(.(x, y)) -> A(x)
   A(b1(x)) -> B1(a(x))
   A(b1(x)) -> A(x)
   F(b(x)) -> F(x)
   .^1(b(x), y) -> .^1(x, y)
   A(f(.(0, x))) -> B1(.(f(.(0, x)), .(0, f(x))))
   A(f(.(0, x))) -> .^1(f(.(0, x)), .(0, f(x)))
   A(f(.(0, x))) -> .^1(0, f(x))
   A(f(.(0, x))) -> F(x)
   A(f(0)) -> B1(.(f(0), 0))
   A(f(0)) -> .^1(f(0), 0)
   F(.(0, x)) -> .^1(0, f(x))
   F(.(0, x)) -> F(x)
   C(b(x)) -> C(a(x))
   C(b(x)) -> A(x)
   A(b(x)) -> A(x)

The TRS R consists of the following rules:

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