/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QReductionProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (41) YES
    (42) QDP
        (43) UsableRulesProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) QReductionProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) NonInfProof [EQUIVALENT, 60 ms]
        (48) QDP
        (49) PisEmptyProof [EQUIVALENT, 0 ms]
        (50) YES


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

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

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MIN(s(x), s(y)) -> MIN(x, y)
   LEN(cons(x, xs)) -> LEN(xs)
   SUM(x, s(y)) -> SUM(x, y)
   LE(s(x), s(y)) -> LE(x, y)
   TAKE(s(x), cons(y, ys)) -> TAKE(x, ys)
   ADDLIST(x, y) -> IF(le(0, min(len(x), len(y))), 0, x, y, nil)
   ADDLIST(x, y) -> LE(0, min(len(x), len(y)))
   ADDLIST(x, y) -> MIN(len(x), len(y))
   ADDLIST(x, y) -> LEN(x)
   ADDLIST(x, y) -> LEN(y)
   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))
   IF(true, c, xs, ys, z) -> LE(s(c), min(len(xs), len(ys)))
   IF(true, c, xs, ys, z) -> MIN(len(xs), len(ys))
   IF(true, c, xs, ys, z) -> LEN(xs)
   IF(true, c, xs, ys, z) -> LEN(ys)
   IF(true, c, xs, ys, z) -> SUM(take(c, xs), take(c, ys))
   IF(true, c, xs, ys, z) -> TAKE(c, xs)
   IF(true, c, xs, ys, z) -> TAKE(c, ys)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 12 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(x), cons(y, ys)) -> TAKE(x, ys)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(x), cons(y, ys)) -> TAKE(x, ys)

R is empty.
The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(x), cons(y, ys)) -> TAKE(x, ys)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*TAKE(s(x), cons(y, ys)) -> TAKE(x, ys)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LE(s(x), s(y)) -> LE(x, y)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LE(s(x), s(y)) -> LE(x, y)

R is empty.
The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LE(s(x), s(y)) -> LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(19) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LE(s(x), s(y)) -> LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(20)
YES

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(x, s(y)) -> SUM(x, y)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(22) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(x, s(y)) -> SUM(x, y)

R is empty.
The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(24) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(x, s(y)) -> SUM(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*SUM(x, s(y)) -> SUM(x, y)
The graph contains the following edges 1 >= 1, 2 > 2


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

(27)
YES

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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LEN(cons(x, xs)) -> LEN(xs)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(29) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LEN(cons(x, xs)) -> LEN(xs)

R is empty.
The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(31) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LEN(cons(x, xs)) -> LEN(xs)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LEN(cons(x, xs)) -> LEN(xs)
The graph contains the following edges 1 > 1


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

(34)
YES

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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MIN(s(x), s(y)) -> MIN(x, y)

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(36) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MIN(s(x), s(y)) -> MIN(x, y)

R is empty.
The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(38) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MIN(s(x), s(y)) -> MIN(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(40) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MIN(s(x), s(y)) -> MIN(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(41)
YES

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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The TRS R consists of the following rules:

   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   addList(x, y) -> if(le(0, min(len(x), len(y))), 0, x, y, nil)
   if(false, c, x, y, z) -> z
   if(true, c, xs, ys, z) -> if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(43) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The TRS R consists of the following rules:

   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))
   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(45) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   addList(x0, x1)
   if(false, x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)


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

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

The TRS R consists of the following rules:

   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(47) NonInfProof (EQUIVALENT)
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:

Note that final constraints are written in bold face.



For Pair IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z)) the following chains were created:
*We consider the chain IF(true, x0, x1, x2, x3) -> IF(le(s(x0), min(len(x1), len(x2))), s(x0), x1, x2, cons(sum(take(x0, x1), take(x0, x2)), x3)), IF(true, x4, x5, x6, x7) -> IF(le(s(x4), min(len(x5), len(x6))), s(x4), x5, x6, cons(sum(take(x4, x5), take(x4, x6)), x7)) which results in the following constraint:

(1)    (IF(le(s(x0), min(len(x1), len(x2))), s(x0), x1, x2, cons(sum(take(x0, x1), take(x0, x2)), x3))=IF(true, x4, x5, x6, x7)  ==>  IF(true, x4, x5, x6, x7)_>=_IF(le(s(x4), min(len(x5), len(x6))), s(x4), x5, x6, cons(sum(take(x4, x5), take(x4, x6)), x7)))



We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:

(2)    (s(x0)=x8 & min(x10, x11)=x9 & le(x8, x9)=true & cons(sum(take(x0, x1), take(x0, x2)), x3)=x7  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x8, x9)=true which results in the following new constraints:

(3)    (le(x14, x13)=true & s(x0)=s(x14) & min(x10, x11)=s(x13) & cons(sum(take(x0, x1), take(x0, x2)), x3)=x7 & (\/x15,x16,x17,x18,x19,x20,x21:le(x14, x13)=true & s(x15)=x14 & min(x16, x17)=x13 & cons(sum(take(x15, x18), take(x15, x19)), x20)=x21  ==>  IF(true, s(x15), x18, x19, x21)_>=_IF(le(s(s(x15)), min(len(x18), len(x19))), s(s(x15)), x18, x19, cons(sum(take(s(x15), x18), take(s(x15), x19)), x21)))  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))

(4)    (true=true & s(x0)=0 & min(x10, x11)=x22 & cons(sum(take(x0, x1), take(x0, x2)), x3)=x7  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))



We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

(5)    (le(x14, x13)=true & x0=x14 & min(x10, x11)=s(x13) & cons(sum(take(x0, x1), take(x0, x2)), x3)=x7 & (\/x15,x16,x17,x18,x19,x20,x21:le(x14, x13)=true & s(x15)=x14 & min(x16, x17)=x13 & cons(sum(take(x15, x18), take(x15, x19)), x20)=x21  ==>  IF(true, s(x15), x18, x19, x21)_>=_IF(le(s(s(x15)), min(len(x18), len(x19))), s(s(x15)), x18, x19, cons(sum(take(s(x15), x18), take(s(x15), x19)), x21)))  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))



We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on min(x10, x11)=s(x13) which results in the following new constraint:

(6)    (min(x26, x25)=s(x13) & le(x14, x13)=true & x0=x14 & cons(sum(take(x0, x1), take(x0, x2)), x3)=x7 & (\/x15,x16,x17,x18,x19,x20,x21:le(x14, x13)=true & s(x15)=x14 & min(x16, x17)=x13 & cons(sum(take(x15, x18), take(x15, x19)), x20)=x21  ==>  IF(true, s(x15), x18, x19, x21)_>=_IF(le(s(s(x15)), min(len(x18), len(x19))), s(s(x15)), x18, x19, cons(sum(take(s(x15), x18), take(s(x15), x19)), x21))) & (\/x27,x28,x29,x30,x31,x32,x33,x34,x35,x36,x37,x38,x39,x40:min(x26, x25)=s(x27) & le(x28, x27)=true & x29=x28 & cons(sum(take(x29, x30), take(x29, x31)), x32)=x33 & (\/x34,x35,x36,x37,x38,x39,x40:le(x28, x27)=true & s(x34)=x28 & min(x35, x36)=x27 & cons(sum(take(x34, x37), take(x34, x38)), x39)=x40  ==>  IF(true, s(x34), x37, x38, x40)_>=_IF(le(s(s(x34)), min(len(x37), len(x38))), s(s(x34)), x37, x38, cons(sum(take(s(x34), x37), take(s(x34), x38)), x40)))  ==>  IF(true, s(x29), x30, x31, x33)_>=_IF(le(s(s(x29)), min(len(x30), len(x31))), s(s(x29)), x30, x31, cons(sum(take(s(x29), x30), take(s(x29), x31)), x33)))  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))



We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (\/x27,x28,x29,x30,x31,x32,x33,x34,x35,x36,x37,x38,x39,x40:min(x26, x25)=s(x27) & le(x28, x27)=true & x29=x28 & cons(sum(take(x29, x30), take(x29, x31)), x32)=x33 & (\/x34,x35,x36,x37,x38,x39,x40:le(x28, x27)=true & s(x34)=x28 & min(x35, x36)=x27 & cons(sum(take(x34, x37), take(x34, x38)), x39)=x40  ==>  IF(true, s(x34), x37, x38, x40)_>=_IF(le(s(s(x34)), min(len(x37), len(x38))), s(s(x34)), x37, x38, cons(sum(take(s(x34), x37), take(s(x34), x38)), x40)))  ==>  IF(true, s(x29), x30, x31, x33)_>=_IF(le(s(s(x29)), min(len(x30), len(x31))), s(s(x29)), x30, x31, cons(sum(take(s(x29), x30), take(s(x29), x31)), x33))) with sigma = [x27 / x13, x28 / x14, x29 / x0, x30 / x1, x31 / x2, x32 / x3, x33 / x7, x34 / x15, x37 / x18, x38 / x19, x40 / x21, x35 / x16, x36 / x17, x39 / x20] which results in the following new constraint:

(7)    (IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7))  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))








To summarize, we get the following constraints P__>=_ for the following pairs.

*IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

*(IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7))  ==>  IF(true, s(x0), x1, x2, x7)_>=_IF(le(s(s(x0)), min(len(x1), len(x2))), s(s(x0)), x1, x2, cons(sum(take(s(x0), x1), take(s(x0), x2)), x7)))








The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 

Using the following integer polynomial ordering the  resulting constraints can be solved 

Polynomial interpretation [NONINF]:

   POL(0) = 1
   POL(IF(x_1, x_2, x_3, x_4, x_5)) = -1 - x_1 - x_2 - x_3 - x_4 - x_5
   POL(c) = -1
   POL(cons(x_1, x_2)) = 1 + x_2
   POL(false) = 1
   POL(le(x_1, x_2)) = 1
   POL(len(x_1)) = 1 + x_1
   POL(min(x_1, x_2)) = 1
   POL(nil) = 1
   POL(s(x_1)) = 1 + x_1
   POL(sum(x_1, x_2)) = 1
   POL(take(x_1, x_2)) = 1 + x_1 + x_2
   POL(true) = 1


The following pairs  are in P_>:
   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))
The following pairs are in P_bound:
   IF(true, c, xs, ys, z) -> IF(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))
The following rules are usable:
   false -> le(s(x), 0)
   le(x, y) -> le(s(x), s(y))
   true -> le(0, x)

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

(48)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   len(nil) -> 0
   len(cons(x, xs)) -> s(len(xs))
   min(0, y) -> 0
   min(s(x), 0) -> 0
   min(s(x), s(y)) -> min(x, y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   take(0, cons(y, ys)) -> y
   take(s(x), cons(y, ys)) -> take(x, ys)
   sum(x, 0) -> x
   sum(x, s(y)) -> s(sum(x, y))
   le(0, x) -> true

The set Q consists of the following terms:

   min(0, x0)
   min(s(x0), 0)
   min(s(x0), s(x1))
   len(nil)
   len(cons(x0, x1))
   sum(x0, 0)
   sum(x0, s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   take(0, cons(x0, x1))
   take(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(49) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(50)
YES