TRS Innermost pair #487524003

details

property	value
status	complete
benchmark	#4.35.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.39358997345 seconds
cpu usage	5.99134541
max memory	4.14990336E8

stage attributes

key	value
output-size	16957
starexec-result	YES

output

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


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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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 29 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 65 ms]
        (14) QDP
        (15) PisEmptyProof [EQUIVALENT, 0 ms]
        (16) YES


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

   and(true, y) -> y
   and(false, y) -> false
   eq(nil, nil) -> true
   eq(cons(t, l), nil) -> false
   eq(nil, cons(t, l)) -> false
   eq(cons(t, l), cons(t', l')) -> and(eq(t, t'), eq(l, l'))
   eq(var(l), var(l')) -> eq(l, l')
   eq(var(l), apply(t, s)) -> false
   eq(var(l), lambda(x, t)) -> false
   eq(apply(t, s), var(l)) -> false
   eq(apply(t, s), apply(t', s')) -> and(eq(t, t'), eq(s, s'))
   eq(apply(t, s), lambda(x, t)) -> false
   eq(lambda(x, t), var(l)) -> false
   eq(lambda(x, t), apply(t, s)) -> false
   eq(lambda(x, t), lambda(x', t')) -> and(eq(x, x'), eq(t, t'))
   if(true, var(k), var(l')) -> var(k)
   if(false, var(k), var(l')) -> var(l')
   ren(var(l), var(k), var(l')) -> if(eq(l, l'), var(k), var(l'))
   ren(x, y, apply(t, s)) -> apply(ren(x, y, t), ren(x, y, s))
   ren(x, y, lambda(z, t)) -> lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

The set Q consists of the following terms:

   and(true, x0)
   and(false, x0)
   eq(nil, nil)
   eq(cons(x0, x1), nil)
   eq(nil, cons(x0, x1))
   eq(cons(x0, x1), cons(x2, x3))
   eq(var(x0), var(x1))
   eq(var(x0), apply(x1, x2))
   eq(var(x0), lambda(x1, x2))
   eq(apply(x0, x1), var(x2))
   eq(apply(x0, x1), apply(x2, x3))
   eq(apply(x0, x1), lambda(x2, x0))
   eq(lambda(x0, x1), var(x2))
   eq(lambda(x0, x1), apply(x1, x2))
   eq(lambda(x0, x1), lambda(x2, x3))
   if(true, var(x0), var(x1))
   if(false, var(x0), var(x1))
   ren(var(x0), var(x1), var(x2))
   ren(x0, x1, apply(x2, x3))
   ren(x0, x1, lambda(x2, x3))


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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(cons(t, l), cons(t', l')) -> AND(eq(t, t'), eq(l, l'))
   EQ(cons(t, l), cons(t', l')) -> EQ(t, t')
   EQ(cons(t, l), cons(t', l')) -> EQ(l, l')
   EQ(var(l), var(l')) -> EQ(l, l')
   EQ(apply(t, s), apply(t', s')) -> AND(eq(t, t'), eq(s, s'))

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