TRS Standard pair #516967450

details

property	value
status	complete
benchmark	secret5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n177.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.65319895744 seconds
cpu usage	6.871316232
max memory	4.99204096E8

stage attributes

key	value
output-size	12660
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 21 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) QDPOrderProof [EQUIVALENT, 68 ms]
        (22) QDP
        (23) DependencyGraphProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES


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

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

   t(N) -> cs(r(q(N)), nt(ns(N)))
   q(0) -> 0
   q(s(X)) -> s(p(q(X), d(X)))
   d(0) -> 0
   d(s(X)) -> s(s(d(X)))
   p(0, X) -> X
   p(X, 0) -> X
   p(s(X), s(Y)) -> s(s(p(X, Y)))
   f(0, X) -> nil
   f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z)))
   t(X) -> nt(X)
   s(X) -> ns(X)
   f(X1, X2) -> nf(X1, X2)
   a(nt(X)) -> t(a(X))
   a(ns(X)) -> s(a(X))
   a(nf(X1, X2)) -> f(a(X1), a(X2))
   a(X) -> 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:

   T(N) -> Q(N)
   Q(s(X)) -> S(p(q(X), d(X)))
   Q(s(X)) -> P(q(X), d(X))
   Q(s(X)) -> Q(X)
   Q(s(X)) -> D(X)
   D(s(X)) -> S(s(d(X)))
   D(s(X)) -> S(d(X))
   D(s(X)) -> D(X)
   P(s(X), s(Y)) -> S(s(p(X, Y)))
   P(s(X), s(Y)) -> S(p(X, Y))
   P(s(X), s(Y)) -> P(X, Y)
   F(s(X), cs(Y, Z)) -> A(Z)
   A(nt(X)) -> T(a(X))
   A(nt(X)) -> A(X)
   A(ns(X)) -> S(a(X))
   A(ns(X)) -> A(X)
   A(nf(X1, X2)) -> F(a(X1), a(X2))
   A(nf(X1, X2)) -> A(X1)

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