YES Termination proof of AG_#3.6_rand.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 RelADPP
        ↳6 RelADPCleverAfsProof (⇒, 123 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 RelADPP
        ↳17 RelADPCleverAfsProof (⇒, 134 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 16 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 RelADPP
        ↳28 RelADPCleverAfsProof (⇒, 128 ms)
        ↳29 QDP
        ↳30 MRRProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 QDPOrderProof (⇔, 17 ms)
        ↳35 QDP
        ↳36 PisEmptyProof (⇔, 0 ms)
        ↳37 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))
rand(x) → x

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → LE(x, y)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → PRED(minus(x, y))
minus(x, s(y)) → pred(MINUS(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
gcd(s(x), s(y)) → if_gcd(LE(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
if_gcd(true, s(x), s(y)) → gcd(MINUS(x, y), s(y))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))
if_gcd(false, s(x), s(y)) → gcd(MINUS(y, x), s(x))

and relative ADPs:

rand(x) → RAND(s(x))
rand(x) → x

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
3 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 3 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, s(y)) → pred(MINUS(x, y))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → x
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
pred(s(x)) → x

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(6) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = MINUS_2 = 0 true = gcd_2 = pred_1 = 0 = le_2 = 0, 1 minus_2 = 1 if_gcd_3 = 0 rand_1 = false =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
MINUS(x1, x2)  =  x2
s(x1)  =  s(x1)
le(x1, x2)  =  le
0  =  0
false  =  false
gcd(x1, x2)  =  gcd(x1, x2)
if_gcd(x1, x2, x3)  =  if_gcd(x2, x3)
true  =  true
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:

0 > [s1, le, false, true] > [gcd2, ifgcd2] > minus1

Status:
s1: multiset
le: []
0: multiset
false: multiset
gcd2: multiset
ifgcd2: multiset
true: multiset
minus1: [1]

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x
rand0(x) → x

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

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

le → false0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(false0) = 0   
POL(gcd0(x1, x2)) = 2·x1 + 2·x2   
POL(if_gcd(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 2   

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

le → le
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x

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

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

gcd0(s0(x), 00) → s0(x)
gcd0(00, y) → y
le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(MINUS(x1)) = x1   
POL(gcd0(x1, x2)) = 2·x1 + 2·x2   
POL(if_gcd(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 1   

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MINUS(s0(y)) → MINUS(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
s0(x1)  =  s0(x1)
le  =  le
if_gcd(x1, x2)  =  if_gcd
gcd0(x1, x2)  =  gcd0
minus(x1)  =  minus(x1)
rand0(x1)  =  rand0
pred0(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[ifgcd, gcd0] > [s01, rand0]
[ifgcd, gcd0] > minus1

Status:
s01: multiset
le: multiset
ifgcd: multiset
gcd0: multiset
minus1: multiset
rand0: []


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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

(13) Obligation:

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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) YES

(16) Obligation:

Relative ADP Problem with
absolute ADPs:

gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → x
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
le(s(x), s(y)) → LE(x, y)
pred(s(x)) → x

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(17) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = true = gcd_2 = LE_2 = 1 0 = le_2 = 0, 1 pred_1 = if_gcd_3 = 0 minus_2 = 1 rand_1 = false =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x1)
s(x1)  =  s(x1)
le(x1, x2)  =  le
0  =  0
false  =  false
gcd(x1, x2)  =  gcd(x1, x2)
if_gcd(x1, x2, x3)  =  if_gcd(x2, x3)
true  =  true
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:

0 > false > s1 > [LE1, minus1]
0 > true > s1 > [LE1, minus1]
[gcd2, ifgcd2] > le > false > s1 > [LE1, minus1]
[gcd2, ifgcd2] > le > true > s1 > [LE1, minus1]

Status:
LE1: multiset
s1: multiset
le: []
0: multiset
false: multiset
gcd2: multiset
ifgcd2: multiset
true: multiset
minus1: [1]

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s0(x)) → LE(x)

The TRS R consists of the following rules:

le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x
rand0(x) → x

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

(19) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

le → false0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(false0) = 0   
POL(gcd0(x1, x2)) = 2·x1 + 2·x2   
POL(if_gcd(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 2   

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s0(x)) → LE(x)

The TRS R consists of the following rules:

le → le
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x

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

(21) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

gcd0(s0(x), 00) → s0(x)
gcd0(00, y) → y
le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(LE(x1)) = x1   
POL(gcd0(x1, x2)) = 2·x1 + 2·x2   
POL(if_gcd(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 1   

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s0(x)) → LE(x)

The TRS R consists of the following rules:

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


LE(s0(x)) → LE(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( le ) = 0

POL( if_gcd(x1, x2) ) = max{0, 2x1 + x2 - 1}

POL( s0(x1) ) = 2x1 + 2

POL( gcd0(x1, x2) ) = 2x1 + x2

POL( minus(x1) ) = x1 + 1

POL( rand0(x1) ) = max{0, -2}

POL( pred0(x1) ) = max{0, x1 - 2}

POL( LE(x1) ) = x1


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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

(24) Obligation:

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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) YES

(27) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(x), s(y)) → le(x, y)
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
le(0, y) → true
minus(x, 0) → x

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(28) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:true = IF_GCD_3 = 0 gcd_2 = pred_1 = if_gcd_3 = 0 rand_1 = false = GCD_2 = s_1 = le_2 = 0, 1 0 = minus_2 = 1
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
IF_GCD(x1, x2, x3)  =  IF_GCD(x2, x3)
false  =  false
s(x1)  =  s(x1)
GCD(x1, x2)  =  GCD(x1, x2)
minus(x1, x2)  =  minus(x1)
le(x1, x2)  =  le
true  =  true
0  =  0
gcd(x1, x2)  =  gcd(x1, x2)
if_gcd(x1, x2, x3)  =  if_gcd(x2, x3)
pred(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:

s1 > [IFGCD2, GCD2] > [false, le, true]
s1 > minus1 > [false, le, true]
s1 > [gcd2, ifgcd2] > [false, le, true]
0 > [false, le, true]

Status:
IFGCD2: multiset
false: multiset
s1: multiset
GCD2: multiset
minus1: multiset
le: []
true: multiset
0: multiset
gcd2: multiset
ifgcd2: multiset

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x
rand0(x) → x

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

(30) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

le → false0
gcd0(s0(x), 00) → s0(x)
gcd0(00, y) → y
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(GCD0(x1, x2)) = 2·x1 + 2·x2   
POL(IF_GCD(x1, x2)) = 2·x1 + 2·x2   
POL(false0) = 0   
POL(gcd0(x1, x2)) = 2 + x1 + x2   
POL(if_gcd(x1, x2)) = 2 + x1 + x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 2   

(31) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
le → true0
minus(x) → x

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

(32) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(GCD0(x1, x2)) = 2·x1 + 2·x2   
POL(IF_GCD(x1, x2)) = 2·x1 + 2·x2   
POL(gcd0(x1, x2)) = x1 + x2   
POL(if_gcd(x1, x2)) = x1 + x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(33) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( le ) = 0

POL( if_gcd(x1, x2) ) = 0

POL( s0(x1) ) = 2x1 + 2

POL( gcd0(x1, x2) ) = 0

POL( minus(x1) ) = x1 + 1

POL( rand0(x1) ) = max{0, -2}

POL( pred0(x1) ) = max{0, x1 - 2}

POL( IF_GCD(x1, x2) ) = 2x1 + x2

POL( GCD0(x1, x2) ) = 2x1 + x2 + 1


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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

(35) Obligation:

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

le → le
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
minus(x) → x

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

(36) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(37) YES