/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 12 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) QDPQMonotonicMRRProof [EQUIVALENT, 37 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) DependencyGraphProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) DependencyGraphProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) TransformationProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) TransformationProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) TransformationProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) UsableRulesProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) TransformationProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) TransformationProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) DependencyGraphProof [EQUIVALENT, 0 ms]
        (60) TRUE


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

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

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)

The TRS R 2 is 
   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   a -> b
   a -> c

The signature Sigma is {gcd_2, gcd2_3, if1_7, pair_2, if2_6, if3_5, if4_4, a, b, c}
----------------------------------------

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

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a


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

(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:

   GCD(x, y) -> GCD2(x, y, 0)
   GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   GCD2(x, y, i) -> LE(x, 0)
   GCD2(x, y, i) -> LE(y, 0)
   GCD2(x, y, i) -> LE(x, y)
   GCD2(x, y, i) -> LE(y, x)
   GCD2(x, y, i) -> INC(i)
   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   IF3(false, b3, x, y, i) -> MINUS(x, y)
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   IF4(false, x, y, i) -> MINUS(y, x)
   INC(s(i)) -> INC(i)
   LE(s(x), s(y)) -> LE(x, y)
   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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:

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

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

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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].

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a


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

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

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

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:
*MINUS(s(x), s(y)) -> MINUS(x, y)
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:

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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].

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a


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

(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:

   INC(s(i)) -> INC(i)

The TRS R consists of the following rules:

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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:

   INC(s(i)) -> INC(i)

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

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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].

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a


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

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

   INC(s(i)) -> INC(i)

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:
*INC(s(i)) -> INC(i)
The graph contains the following edges 1 > 1


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

(27)
YES

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

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

   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)

The TRS R consists of the following rules:

   gcd(x, y) -> gcd2(x, y, 0)
   gcd2(x, y, i) -> if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   if1(true, b1, b2, b3, x, y, i) -> pair(result(y), neededIterations(i))
   if1(false, b1, b2, b3, x, y, i) -> if2(b1, b2, b3, x, y, i)
   if2(true, b2, b3, x, y, i) -> pair(result(x), neededIterations(i))
   if2(false, b2, b3, x, y, i) -> if3(b2, b3, x, y, i)
   if3(false, b3, x, y, i) -> gcd2(minus(x, y), y, i)
   if3(true, b3, x, y, i) -> if4(b3, x, y, i)
   if4(false, x, y, i) -> gcd2(x, minus(y, x), i)
   if4(true, x, y, i) -> pair(result(x), neededIterations(i))
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   a -> b
   a -> c

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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:

   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))
   a

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].

   gcd(x0, x1)
   gcd2(x0, x1, x2)
   if1(true, x0, x1, x2, x3, x4, x5)
   if1(false, x0, x1, x2, x3, x4, x5)
   if2(true, x0, x1, x2, x3, x4)
   if2(false, x0, x1, x2, x3, x4)
   if3(false, x0, x1, x2, x3)
   if3(true, x0, x1, x2, x3)
   if4(false, x0, x1, x2)
   if4(true, x0, x1, x2)
   a


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

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

   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(33) QDPQMonotonicMRRProof (EQUIVALENT)
By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.


Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(GCD2(x_1, x_2, x_3)) = x_1 + 2*x_2
   POL(IF1(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_5 + 2*x_6
   POL(IF2(x_1, x_2, x_3, x_4, x_5, x_6)) = x_4 + 2*x_5
   POL(IF3(x_1, x_2, x_3, x_4, x_5)) = x_3 + 2*x_4
   POL(IF4(x_1, x_2, x_3, x_4)) = x_2 + 2*x_3
   POL(false) = 0
   POL(inc(x_1)) = x_1
   POL(le(x_1, x_2)) = 0
   POL(minus(x_1, x_2)) = x_1
   POL(s(x_1)) = 2 + 2*x_1
   POL(true) = 0


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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(35) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule GCD2(x, y, i) -> IF1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i)) at position [0] we obtained the following new rules [LPAR04]:

   (GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2)),GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2)))
   (GCD2(0, y1, y2) -> IF1(true, le(y1, 0), le(0, y1), le(y1, 0), 0, y1, inc(y2)),GCD2(0, y1, y2) -> IF1(true, le(y1, 0), le(0, y1), le(y1, 0), 0, y1, inc(y2)))


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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   GCD2(0, y1, y2) -> IF1(true, le(y1, 0), le(0, y1), le(y1, 0), 0, y1, inc(y2))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(37) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(39) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF3(false, b3, x, y, i) -> GCD2(minus(x, y), y, i) at position [0] we obtained the following new rules [LPAR04]:

   (IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3),IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3))
   (IF3(false, y0, 0, x0, y3) -> GCD2(0, x0, y3),IF3(false, y0, 0, x0, y3) -> GCD2(0, x0, y3))


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF3(false, y0, 0, x0, y3) -> GCD2(0, x0, y3)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(41) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

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

   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i)
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(43) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF1(false, b1, b2, b3, x, y, i) -> IF2(b1, b2, b3, x, y, i) we obtained the following new rules [LPAR04]:

   (IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3),IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3))


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

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

   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i)
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(45) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF2(false, b2, b3, x, y, i) -> IF3(b2, b3, x, y, i) we obtained the following new rules [LPAR04]:

   (IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5),IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5))


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

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

   IF3(true, b3, x, y, i) -> IF4(b3, x, y, i)
   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(47) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF3(true, b3, x, y, i) -> IF4(b3, x, y, i) we obtained the following new rules [LPAR04]:

   (IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4),IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4))


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

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

   IF4(false, x, y, i) -> GCD2(x, minus(y, x), i)
   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(49) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF4(false, x, y, i) -> GCD2(x, minus(y, x), i) we obtained the following new rules [LPAR04]:

   (IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3),IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3))


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

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)
   IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(51) 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.
----------------------------------------

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3)
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)
   IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(53) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF3(false, y0, x0, 0, y3) -> GCD2(x0, 0, y3) we obtained the following new rules [LPAR04]:

   (IF3(false, z1, s(z2), 0, z4) -> GCD2(s(z2), 0, z4),IF3(false, z1, s(z2), 0, z4) -> GCD2(s(z2), 0, z4))


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

(54)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3)
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)
   IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3)
   IF3(false, z1, s(z2), 0, z4) -> GCD2(s(z2), 0, z4)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(55) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF1(false, y_0, y_1, y_2, s(z0), z1, y_3) -> IF2(y_0, y_1, y_2, s(z0), z1, y_3) we obtained the following new rules [LPAR04]:

   (IF1(false, false, x1, x2, s(x3), x4, x5) -> IF2(false, x1, x2, s(x3), x4, x5),IF1(false, false, x1, x2, s(x3), x4, x5) -> IF2(false, x1, x2, s(x3), x4, x5))


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

(56)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2))
   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)
   IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3)
   IF3(false, z1, s(z2), 0, z4) -> GCD2(s(z2), 0, z4)
   IF1(false, false, x1, x2, s(x3), x4, x5) -> IF2(false, x1, x2, s(x3), x4, x5)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(57) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule GCD2(s(x0), y1, y2) -> IF1(false, le(y1, 0), le(s(x0), y1), le(y1, s(x0)), s(x0), y1, inc(y2)) at position [1] we obtained the following new rules [LPAR04]:

   (GCD2(s(y0), s(x0), y2) -> IF1(false, false, le(s(y0), s(x0)), le(s(x0), s(y0)), s(y0), s(x0), inc(y2)),GCD2(s(y0), s(x0), y2) -> IF1(false, false, le(s(y0), s(x0)), le(s(x0), s(y0)), s(y0), s(x0), inc(y2)))
   (GCD2(s(y0), 0, y2) -> IF1(false, true, le(s(y0), 0), le(0, s(y0)), s(y0), 0, inc(y2)),GCD2(s(y0), 0, y2) -> IF1(false, true, le(s(y0), 0), le(0, s(y0)), s(y0), 0, inc(y2)))


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

(58)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, z1, z2, s(z3), z4, z5) -> IF3(z1, z2, s(z3), z4, z5)
   IF3(true, z1, s(z2), z3, z4) -> IF4(z1, s(z2), z3, z4)
   IF4(false, s(z1), z2, z3) -> GCD2(s(z1), minus(z2, s(z1)), z3)
   IF3(false, z1, s(z2), 0, z4) -> GCD2(s(z2), 0, z4)
   IF1(false, false, x1, x2, s(x3), x4, x5) -> IF2(false, x1, x2, s(x3), x4, x5)
   GCD2(s(y0), s(x0), y2) -> IF1(false, false, le(s(y0), s(x0)), le(s(x0), s(y0)), s(y0), s(x0), inc(y2))
   GCD2(s(y0), 0, y2) -> IF1(false, true, le(s(y0), 0), le(0, s(y0)), s(y0), 0, inc(y2))

The TRS R consists of the following rules:

   minus(0, y) -> 0
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(i)) -> s(inc(i))

The set Q consists of the following terms:

   inc(0)
   inc(s(x0))
   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(0, x0)
   minus(s(x0), s(x1))

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

(59) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 7 less nodes.
----------------------------------------

(60)
TRUE