/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) 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) TransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) TransformationProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) UsableRulesProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) DependencyGraphProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) UsableRulesProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) QDPOrderProof [EQUIVALENT, 23 ms]
        (44) QDP
        (45) DependencyGraphProof [EQUIVALENT, 0 ms]
        (46) TRUE


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

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

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(plus, dummy, x, y) -> function(if, function(iszero, x, x, x), x, y)
   function(if, true, x, y) -> y
   function(if, false, x, y) -> function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   function(third, x, y, z) -> 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:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(plus, dummy, x, y) -> function(if, function(iszero, x, x, x), x, y)
   function(if, true, x, y) -> y
   function(if, false, x, y) -> function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   function(third, x, y, z) -> z

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)


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

(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:

   FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)
   FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)
   FUNCTION(plus, dummy, x, y) -> FUNCTION(iszero, x, x, x)
   FUNCTION(if, false, x, y) -> FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   FUNCTION(if, false, x, y) -> FUNCTION(third, x, y, y)
   FUNCTION(if, false, x, y) -> FUNCTION(p, x, x, y)

The TRS R consists of the following rules:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(plus, dummy, x, y) -> function(if, function(iszero, x, x, x), x, y)
   function(if, true, x, y) -> y
   function(if, false, x, y) -> function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   function(third, x, y, z) -> z

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)

The TRS R consists of the following rules:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(plus, dummy, x, y) -> function(if, function(iszero, x, x, x), x, y)
   function(if, true, x, y) -> y
   function(if, false, x, y) -> function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   function(third, x, y, z) -> z

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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:

   FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)

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

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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].

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)


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

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

   FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)

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:
*FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)
The graph contains the following edges 1 >= 1, 2 > 2, 2 > 3, 2 > 4


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

(13)
YES

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

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

   FUNCTION(if, false, x, y) -> FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)

The TRS R consists of the following rules:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(plus, dummy, x, y) -> function(if, function(iszero, x, x, x), x, y)
   function(if, true, x, y) -> y
   function(if, false, x, y) -> function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   function(third, x, y, z) -> z

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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:

   FUNCTION(if, false, x, y) -> FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
   FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)

The TRS R consists of the following rules:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(third, x, y, z) -> z
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(17) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule FUNCTION(if, false, x, y) -> FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y)) at position [1] we obtained the following new rules [LPAR04]:

   (FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y)),FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y)))


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

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

   FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)
   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

The TRS R consists of the following rules:

   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false
   function(third, x, y, z) -> z
   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(19) 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.
----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)
   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(21) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y) we obtained the following new rules [LPAR04]:

   (FUNCTION(plus, z1, y_0, s(z1)) -> FUNCTION(if, function(iszero, y_0, y_0, y_0), y_0, s(z1)),FUNCTION(plus, z1, y_0, s(z1)) -> FUNCTION(if, function(iszero, y_0, y_0, y_0), y_0, s(z1)))


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))
   FUNCTION(plus, z1, y_0, s(z1)) -> FUNCTION(if, function(iszero, y_0, y_0, y_0), y_0, s(z1))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(23) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule FUNCTION(plus, z1, y_0, s(z1)) -> FUNCTION(if, function(iszero, y_0, y_0, y_0), y_0, s(z1)) at position [1] we obtained the following new rules [LPAR04]:

   (FUNCTION(plus, y0, 0, s(y0)) -> FUNCTION(if, true, 0, s(y0)),FUNCTION(plus, y0, 0, s(y0)) -> FUNCTION(if, true, 0, s(y0)))
   (FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0)),FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0)))


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))
   FUNCTION(plus, y0, 0, s(y0)) -> FUNCTION(if, true, 0, s(y0))
   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))
   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))
   function(iszero, 0, dummy, dummy2) -> true
   function(iszero, s(x), dummy, dummy2) -> false

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(27) 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.
----------------------------------------

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

   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))
   FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(29) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y)) at position [2] we obtained the following new rules [LPAR04]:

   (FUNCTION(if, false, 0, x1) -> FUNCTION(plus, x1, 0, s(x1)),FUNCTION(if, false, 0, x1) -> FUNCTION(plus, x1, 0, s(x1)))
   (FUNCTION(if, false, s(0), x1) -> FUNCTION(plus, x1, 0, s(x1)),FUNCTION(if, false, s(0), x1) -> FUNCTION(plus, x1, 0, s(x1)))
   (FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2)),FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2)))


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

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

   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))
   FUNCTION(if, false, 0, x1) -> FUNCTION(plus, x1, 0, s(x1))
   FUNCTION(if, false, s(0), x1) -> FUNCTION(plus, x1, 0, s(x1))
   FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(31) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2))
   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))

The TRS R consists of the following rules:

   function(p, 0, dummy, dummy2) -> 0
   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(33) 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.
----------------------------------------

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

   FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2))
   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(35) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FUNCTION(if, false, s(s(x0)), x2) -> FUNCTION(plus, x2, s(function(p, s(x0), x0, x0)), s(x2)) we obtained the following new rules [LPAR04]:

   (FUNCTION(if, false, s(s(x0)), s(z0)) -> FUNCTION(plus, s(z0), s(function(p, s(x0), x0, x0)), s(s(z0))),FUNCTION(if, false, s(s(x0)), s(z0)) -> FUNCTION(plus, s(z0), s(function(p, s(x0), x0, x0)), s(s(z0))))


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

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

   FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0))
   FUNCTION(if, false, s(s(x0)), s(z0)) -> FUNCTION(plus, s(z0), s(function(p, s(x0), x0, x0)), s(s(z0)))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(37) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FUNCTION(plus, y0, s(x0), s(y0)) -> FUNCTION(if, false, s(x0), s(y0)) we obtained the following new rules [LPAR04]:

   (FUNCTION(plus, s(z1), s(y_0), s(s(z1))) -> FUNCTION(if, false, s(y_0), s(s(z1))),FUNCTION(plus, s(z1), s(y_0), s(s(z1))) -> FUNCTION(if, false, s(y_0), s(s(z1))))


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

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

   FUNCTION(if, false, s(s(x0)), s(z0)) -> FUNCTION(plus, s(z0), s(function(p, s(x0), x0, x0)), s(s(z0)))
   FUNCTION(plus, s(z1), s(y_0), s(s(z1))) -> FUNCTION(if, false, s(y_0), s(s(z1)))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(39) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FUNCTION(if, false, s(s(x0)), s(z0)) -> FUNCTION(plus, s(z0), s(function(p, s(x0), x0, x0)), s(s(z0))) we obtained the following new rules [LPAR04]:

   (FUNCTION(if, false, s(s(x0)), s(s(z0))) -> FUNCTION(plus, s(s(z0)), s(function(p, s(x0), x0, x0)), s(s(s(z0)))),FUNCTION(if, false, s(s(x0)), s(s(z0))) -> FUNCTION(plus, s(s(z0)), s(function(p, s(x0), x0, x0)), s(s(s(z0)))))


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

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

   FUNCTION(plus, s(z1), s(y_0), s(s(z1))) -> FUNCTION(if, false, s(y_0), s(s(z1)))
   FUNCTION(if, false, s(s(x0)), s(s(z0))) -> FUNCTION(plus, s(s(z0)), s(function(p, s(x0), x0, x0)), s(s(s(z0))))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(41) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule FUNCTION(plus, s(z1), s(y_0), s(s(z1))) -> FUNCTION(if, false, s(y_0), s(s(z1))) we obtained the following new rules [LPAR04]:

   (FUNCTION(plus, s(x0), s(s(y_0)), s(s(x0))) -> FUNCTION(if, false, s(s(y_0)), s(s(x0))),FUNCTION(plus, s(x0), s(s(y_0)), s(s(x0))) -> FUNCTION(if, false, s(s(y_0)), s(s(x0))))


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

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

   FUNCTION(if, false, s(s(x0)), s(s(z0))) -> FUNCTION(plus, s(s(z0)), s(function(p, s(x0), x0, x0)), s(s(s(z0))))
   FUNCTION(plus, s(x0), s(s(y_0)), s(s(x0))) -> FUNCTION(if, false, s(s(y_0)), s(s(x0)))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(43) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   FUNCTION(if, false, s(s(x0)), s(s(z0))) -> FUNCTION(plus, s(s(z0)), s(function(p, s(x0), x0, x0)), s(s(s(z0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( FUNCTION_4(x_1, ..., x_4) ) = max{0, 2x_3 - 2}
POL( s_1(x_1) ) = x_1 + 2
POL( function_4(x_1, ..., x_4) ) = max{0, 2x_1 + x_2 - 2}
POL( p ) = 0
POL( 0 ) = 0
POL( if ) = 2
POL( false ) = 2
POL( plus ) = 2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))


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

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

   FUNCTION(plus, s(x0), s(s(y_0)), s(s(x0))) -> FUNCTION(if, false, s(s(y_0)), s(s(x0)))

The TRS R consists of the following rules:

   function(p, s(0), dummy, dummy2) -> 0
   function(p, s(s(x)), dummy, dummy2) -> s(function(p, s(x), x, x))

The set Q consists of the following terms:

   function(iszero, 0, x0, x1)
   function(iszero, s(x0), x1, x2)
   function(p, 0, x0, x1)
   function(p, s(0), x0, x1)
   function(p, s(s(x0)), x1, x2)
   function(plus, x0, x1, x2)
   function(if, true, x0, x1)
   function(if, false, x0, x1)
   function(third, x0, x1, x2)

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

(45) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(46)
TRUE