TRS Standard pair #516963970

details

property	value
status	complete
benchmark	IJCAR_18.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n089.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.64194202423 seconds
cpu usage	7.07927815
max memory	5.13470464E8

stage attributes

key	value
output-size	17149
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 2 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, 7 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) DependencyGraphProof [EQUIVALENT, 0 ms]
        (28) TRUE
    (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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:

   PLUS(s(x), y) -> PLUS(x, y)
   TIMES(s(x), y) -> PLUS(y, times(x, y))
   TIMES(s(x), y) -> TIMES(x, y)
   DIV(x, y) -> QUOT(x, y, y)
   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
   QUOT(x, 0, s(z)) -> DIV(x, s(z))
   DIV(div(x, y), z) -> DIV(x, times(y, z))
   DIV(div(x, y), z) -> TIMES(y, z)
   EQ(s(x), s(y)) -> EQ(x, y)
   DIVIDES(y, x) -> EQ(x, times(div(x, y), y))

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