TRS Standard pair #516967815

details

property	value
status	complete
benchmark	intersect.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.3356089592 seconds
cpu usage	6.212124955
max memory	4.7456256E8

stage attributes

key	value
output-size	13091
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 31 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(0, 0) -> true
   eq(0, s(x)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   app(app(l1, l2), l3) -> app(l1, app(l2, l3))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
   ifmem(true, x, l) -> true
   ifmem(false, x, l) -> mem(x, l)
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

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:

   EQ(s(x), s(y)) -> EQ(x, y)
   APP(cons(x, l1), l2) -> APP(l1, l2)
   APP(app(l1, l2), l3) -> APP(l1, app(l2, l3))
   APP(app(l1, l2), l3) -> APP(l2, l3)
   MEM(x, cons(y, l)) -> IFMEM(eq(x, y), x, l)
   MEM(x, cons(y, l)) -> EQ(x, y)
   IFMEM(false, x, l) -> MEM(x, l)
   INTER(app(l1, l2), l3) -> APP(inter(l1, l3), inter(l2, l3))
   INTER(app(l1, l2), l3) -> INTER(l1, l3)
   INTER(app(l1, l2), l3) -> INTER(l2, l3)
   INTER(l1, app(l2, l3)) -> APP(inter(l1, l2), inter(l1, l3))
   INTER(l1, app(l2, l3)) -> INTER(l1, l2)
   INTER(l1, app(l2, l3)) -> INTER(l1, l3)
   INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
   INTER(cons(x, l1), l2) -> MEM(x, l2)
   INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
   INTER(l1, cons(x, l2)) -> MEM(x, l1)
   IFINTER(true, x, l1, l2) -> INTER(l1, l2)
   IFINTER(false, x, l1, l2) -> INTER(l1, l2)

The TRS R consists of the following rules:

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