TRS Standard pair #516965800

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	n059.star.cs.uiowa.edu
space	Hydras

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	21.5289449692 seconds
cpu usage	65.952731598
max memory	3.421437952E9

stage attributes

key	value
output-size	18458
starexec-result	YES

output

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


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 126 ms]
        (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 42 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 31 ms]
        (28) QDP
        (29) QDPOrderProof [EQUIVALENT, 2575 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)

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