YES Termination proof of INVY_15_3.5b_rand.ari

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 (⇒, 141 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 8 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 15 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 RelADPP
        ↳19 RelADPCleverAfsProof (⇒, 137 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 MRRProof (⇔, 7 ms)
        ↳24 QDP
        ↳25 MRRProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔, 15 ms)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔, 0 ms)
        ↳30 YES
    ↳31 RelADPP
        ↳32 RelADPCleverAfsProof (⇒, 134 ms)
        ↳33 QDP
        ↳34 MRRProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 MRRProof (⇔, 7 ms)
        ↳37 QDP
        ↳38 QDPOrderProof (⇔, 16 ms)
        ↳39 QDP
        ↳40 DependencyGraphProof (⇔, 0 ms)
        ↳41 TRUE

(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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → 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)
minus(0, y) → 0
minus(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
minus(s(x), y) → if_minus(LE(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(MINUS(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
mod(s(x), s(y)) → if_mod(LE(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → MOD(minus(x, y), s(y))
if_mod(true, s(x), s(y)) → mod(MINUS(x, y), s(y))
if_mod(false, s(x), s(y)) → 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:

le(s(x), s(y)) → LE(x, y)

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
if_mod(false, s(x), s(y)) → s(x)
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
minus(0, y) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
le(0, y) → true
mod(s(x), 0) → 0
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 = mod_2 = if_mod_3 = 0 true = LE_2 = 0 le_2 = 0, 1 0 = minus_2 = 1 if_minus_3 = 0, 2 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(x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
mod(x1, x2)  =  mod(x1, x2)
if_mod(x1, x2, x3)  =  if_mod(x2, x3)
true  =  true
0  =  0
false  =  false

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

le > [LE1, s1, false] > 0 > true
[mod2, ifmod2] > [LE1, s1, false] > 0 > true

Status:
LE1: multiset
s1: [1]
le: []
mod2: [1,2]
ifmod2: [1,2]
true: multiset
0: multiset
false: multiset

(7) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(s0(x), s0(y)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
mod0(00, y) → 00
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))
le → true0
mod0(s0(x), 00) → 00
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 → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(false0) = 2   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod0(x1, x2)) = 2·x1 + 2·x2   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(9) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(s0(x), s0(y)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
mod0(00, y) → 00
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))
mod0(s0(x), 00) → 00

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:

le → false0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(false0) = 1   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = 2·x1 + 2·x2   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod0(x1, x2)) = 2·x1 + 2·x2   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(11) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
if_mod(s0(x), s0(y)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
mod0(00, y) → 00
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))
mod0(s0(x), 00) → 00

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

(12) 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:

if_mod(s0(x), s0(y)) → s0(x)
mod0(00, y) → 00
mod0(s0(x), 00) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = 1 + x1 + x2   
POL(le) = 0   
POL(minus(x1)) = x1   
POL(mod0(x1, x2)) = 1 + x1 + x2   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(13) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


LE(s0(y)) → LE(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1)  =  LE(x1)
s0(x1)  =  s0(x1)
minus(x1)  =  minus(x1)
if_minus(x1)  =  if_minus(x1)
le  =  le
rand0(x1)  =  rand0
00  =  00
mod0(x1, x2)  =  x1
if_mod(x1, x2)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
LE1 > [s01, minus1, ifminus1, 00]
le > [s01, minus1, ifminus1, 00]
rand0 > [s01, minus1, ifminus1, 00]

Status:
LE1: [1]
s01: multiset
minus1: multiset
ifminus1: multiset
le: multiset
rand0: multiset
00: multiset


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

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))

(15) Obligation:

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

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod0(s0(x), s0(y)) → if_mod(s0(x), s0(y))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x), s0(y)) → mod0(minus(x), s0(y))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) YES

(18) Obligation:

Relative ADP Problem with
absolute ADPs:

if_minus(false, s(x), y) → s(MINUS(x, y))
minus(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
if_mod(false, s(x), s(y)) → s(x)
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
minus(0, y) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
mod(0, y) → 0
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
le(0, y) → true
mod(s(x), 0) → 0
rand(x) → x

(19) 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 = mod_2 = 1 if_mod_3 = 2 true = le_2 = 0, 1 0 = minus_2 = 1 IF_MINUS_3 = 0 if_minus_3 = 0, 2 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.
IF_MINUS(x1, x2, x3)  =  IF_MINUS(x2, x3)
false  =  false
s(x1)  =  s(x1)
MINUS(x1, x2)  =  MINUS(x1, x2)
le(x1, x2)  =  le
0  =  0
true  =  true
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2
mod(x1, x2)  =  mod(x1)
if_mod(x1, x2, x3)  =  if_mod(x1, x2)

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

[s1, le, true] > mod1 > 0 > [IFMINUS2, false, MINUS2, ifmod2]

Status:
IFMINUS2: [1,2]
false: multiset
s1: [1]
MINUS2: [1,2]
le: []
0: multiset
true: multiset
mod1: multiset
ifmod2: multiset

(20) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(false0, s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
mod(00) → 00
if_mod(true0, s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
rand0(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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MINUS(x1, x2)) = x1 + x2   
POL(MINUS0(x1, x2)) = x1 + x2   
POL(false0) = 0   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = x1 + 2·x2   
POL(le) = 0   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2·x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(22) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(false0, s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
mod(00) → 00
if_mod(true0, s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00

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

(23) 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
mod(00) → 00
mod(s0(x)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MINUS(x1, x2)) = x1 + x2   
POL(MINUS0(x1, x2)) = x1 + x2   
POL(false0) = 0   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = 2·x1 + x2   
POL(le) = 1   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 1   

(24) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
if_mod(false0, s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(true0, s0(x)) → mod(minus(x))
le → true0

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

(25) 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:

if_mod(false0, s0(x)) → s0(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MINUS(x1, x2)) = x1 + x2   
POL(MINUS0(x1, x2)) = x1 + x2   
POL(false0) = 2   
POL(if_minus(x1)) = x1   
POL(if_mod(x1, x2)) = x1 + 2·x2   
POL(le) = 0   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2·x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(26) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(true0, s0(x)) → mod(minus(x))
le → true0

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

(27) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( minus(x1) ) = 2x1 + 1

POL( s0(x1) ) = 2x1 + 1

POL( if_minus(x1) ) = 2x1 + 1

POL( le ) = 0

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

POL( 00 ) = 0

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

POL( if_mod(x1, x2) ) = max{0, -2}

POL( true0 ) = 0

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

POL( MINUS0(x1, x2) ) = 2x1 + x2 + 2


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

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(true0, s0(x)) → mod(minus(x))
le → true0

(28) Obligation:

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

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(le, s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(true0, s0(x)) → mod(minus(x))
le → true0

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) YES

(31) Obligation:

Relative ADP Problem with
absolute ADPs:

if_mod(true, s(x), s(y)) → MOD(minus(x, y), s(y))
mod(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
if_mod(false, s(x), s(y)) → s(x)
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
minus(0, y) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
le(0, y) → true
mod(s(x), 0) → 0
rand(x) → x

(32) 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 = mod_2 = 1 if_mod_3 = 0, 2 true = le_2 = 0, 1 0 = MOD_2 = minus_2 = 1 IF_MOD_3 = if_minus_3 = 0, 2 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.
MOD(x1, x2)  =  MOD(x1, x2)
s(x1)  =  s(x1)
IF_MOD(x1, x2, x3)  =  IF_MOD(x1, x2, x3)
le(x1, x2)  =  le
true  =  true
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2
0  =  0
false  =  false
mod(x1, x2)  =  mod(x1)
if_mod(x1, x2, x3)  =  if_mod(x2)

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

s1 > 0 > [MOD2, IFMOD3, le, true, false]
s1 > [mod1, ifmod1] > [MOD2, IFMOD3, le, true, false]

Status:
MOD2: [1,2]
s1: multiset
IFMOD3: [2,1,3]
le: multiset
true: multiset
0: multiset
false: multiset
mod1: [1]
ifmod1: [1]

(33) Obligation:

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

MOD0(s0(x), s0(y)) → IF_MOD0(le, s0(x), s0(y))
IF_MOD0(true0, s0(x), s0(y)) → MOD0(minus(x), s0(y))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(s0(x))
if_minus(s0(x)) → s0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
rand0(x) → x

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

(34) 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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MOD0(x1, x2, x3)) = x1 + x2 + 2·x3   
POL(MOD0(x1, x2)) = x1 + 2·x2   
POL(false0) = 0   
POL(if_minus(x1)) = x1   
POL(if_mod(x1)) = 2·x1   
POL(le) = 0   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2·x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(35) Obligation:

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

MOD0(s0(x), s0(y)) → IF_MOD0(le, s0(x), s0(y))
IF_MOD0(true0, s0(x), s0(y)) → MOD0(minus(x), s0(y))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
le → false0
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(s0(x))
if_minus(s0(x)) → s0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00

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

(36) 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
if_mod(s0(x)) → s0(x)
mod(00) → 00
mod(s0(x)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MOD0(x1, x2, x3)) = 2·x1 + x2 + 2·x3   
POL(MOD0(x1, x2)) = 2 + x1 + 2·x2   
POL(false0) = 0   
POL(if_minus(x1)) = x1   
POL(if_mod(x1)) = 1 + 2·x1   
POL(le) = 1   
POL(minus(x1)) = x1   
POL(mod(x1)) = 1 + 2·x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 1   

(37) Obligation:

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

MOD0(s0(x), s0(y)) → IF_MOD0(le, s0(x), s0(y))
IF_MOD0(true0, s0(x), s0(y)) → MOD0(minus(x), s0(y))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0

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

(38) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF_MOD0(true0, s0(x), s0(y)) → MOD0(minus(x), s0(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MOD0(x1, x2)  =  MOD0(x1)
s0(x1)  =  s0(x1)
IF_MOD0(x1, x2, x3)  =  IF_MOD0(x2)
le  =  le
true0  =  true0
minus(x1)  =  x1
if_minus(x1)  =  x1
rand0(x1)  =  rand0
00  =  00
mod(x1)  =  mod(x1)
if_mod(x1)  =  if_mod(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
rand0 > [MOD01, s01, IFMOD01, mod1, ifmod1] > [le, true0] > 00

Status:
MOD01: multiset
s01: multiset
IFMOD01: multiset
le: multiset
true0: multiset
rand0: multiset
00: multiset
mod1: [1]
ifmod1: [1]


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

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0

(39) Obligation:

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

MOD0(s0(x), s0(y)) → IF_MOD0(le, s0(x), s0(y))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
mod(s0(x)) → if_mod(s0(x))
if_minus(s0(x)) → s0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0

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

(40) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(41) TRUE