YES Termination proof of AG_#3.5a_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 (⇒, 110 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 (⇒, 104 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 56 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 6 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 RelADPP
        ↳28 RelADPCleverAfsProof (⇒, 96 ms)
        ↳29 QDP
        ↳30 MRRProof (⇔, 48 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 QDPOrderProof (⇔, 11 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))
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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → PRED(minus(x, y))
minus(x, s(y)) → pred(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:

minus(x, s(y)) → pred(MINUS(x, y))
mod(s(x), s(y)) → if_mod(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
mod(0, y) → 0
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)
le(0, y) → true
mod(s(x), 0) → 0
minus(x, 0) → 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 mod_2 = 1 if_mod_3 = 0, 2 true = pred_1 = le_2 = 0, 1 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.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)
le(x1, x2)  =  le
0  =  0
false  =  false
if_mod(x1, x2, x3)  =  if_mod(x2)
pred(x1)  =  x1
mod(x1, x2)  =  mod(x1)
minus(x1, x2)  =  minus(x1)
true  =  true

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

[ifmod1, mod1] > 0 > [MINUS1, s1, le, false, minus1] > true

Status:
MINUS1: multiset
s1: multiset
le: multiset
0: multiset
false: multiset
ifmod1: multiset
mod1: multiset
minus1: multiset
true: multiset

(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
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
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
if_mod(s0(x)) → s0(x)
mod(00) → 00
mod(s0(x)) → 00
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(false0) = 0   
POL(if_mod(x1)) = 2 + 2·x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + 2·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
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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:

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(MINUS(x1)) = x1   
POL(if_mod(x1)) = 2 + x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + 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
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( le ) = 0

POL( rand0(x1) ) = 2

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

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

POL( mod(x1) ) = max{0, 2x1 - 1}

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

POL( minus(x1) ) = x1

POL( MINUS(x1) ) = x1


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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(minus(x))
minus(x) → x

(13) Obligation:

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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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:

mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
le(s(x), 0) → false
mod(0, y) → 0
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)
le(0, y) → true
mod(s(x), 0) → 0
minus(x, 0) → 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 = mod_2 = 1 if_mod_3 = 0, 2 true = LE_2 = 1 le_2 = 0, 1 pred_1 = 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
if_mod(x1, x2, x3)  =  x2
pred(x1)  =  x1
mod(x1, x2)  =  x1
minus(x1, x2)  =  minus(x1)
true  =  true

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

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

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

(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
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
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
if_mod(s0(x)) → s0(x)
mod(00) → 00
mod(s0(x)) → 00
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(false0) = 0   
POL(if_mod(x1)) = 2 + 2·x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + 2·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
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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:

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(LE(x1)) = x1   
POL(if_mod(x1)) = 2 + x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + 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
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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: Combined order from the following AFS and order.
LE(x1)  =  x1
s0(x1)  =  s0(x1)
le  =  le
rand0(x1)  =  rand0
pred0(x1)  =  x1
mod(x1)  =  x1
if_mod(x1)  =  x1
minus(x1)  =  minus(x1)

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

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


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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(minus(x))
minus(x) → x

(24) Obligation:

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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(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)
le(s(x), 0) → false
if_mod(false, s(x), s(y)) → s(x)
mod(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
pred(s(x)) → x
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
if_mod(true, s(x), s(y)) → MOD(minus(x, y), 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
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:if_mod_3 = 0, 2 true = pred_1 = IF_MOD_3 = 2 rand_1 = false = s_1 = mod_2 = 1 le_2 = 0, 1 0 = minus_2 = 1 MOD_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.
MOD(x1, x2)  =  MOD(x1)
s(x1)  =  s(x1)
IF_MOD(x1, x2, x3)  =  IF_MOD(x1, x2)
le(x1, x2)  =  le
true  =  true
minus(x1, x2)  =  x1
0  =  0
false  =  false
if_mod(x1, x2, x3)  =  if_mod(x2)
pred(x1)  =  x1
mod(x1, x2)  =  mod(x1)

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

[MOD1, s1] > IFMOD2 > [le, true, false]
[MOD1, s1] > 0 > [le, true, false]
[MOD1, s1] > mod1 > ifmod1 > [le, true, false]

Status:
MOD1: multiset
s1: multiset
IFMOD2: multiset
le: multiset
true: multiset
0: multiset
false: multiset
ifmod1: multiset
mod1: [1]

(29) Obligation:

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

MOD(s0(x)) → IF_MOD(le, s0(x))
IF_MOD(true0, s0(x)) → MOD(minus(x))

The TRS R consists of the following rules:

le → le
le → false0
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MOD(x1, x2)) = x1 + x2   
POL(MOD(x1)) = x1   
POL(false0) = 0   
POL(if_mod(x1)) = 2·x1   
POL(le) = 0   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2·x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(31) Obligation:

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

MOD(s0(x)) → IF_MOD(le, s0(x))
IF_MOD(true0, s0(x)) → MOD(minus(x))

The TRS R consists of the following rules:

le → le
le → false0
if_mod(s0(x)) → s0(x)
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
mod(00) → 00
if_mod(s0(x)) → mod(minus(x))
le → true0
mod(s0(x)) → 00
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 → false0
if_mod(s0(x)) → s0(x)
mod(00) → 00
mod(s0(x)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MOD(x1, x2)) = x1 + x2   
POL(MOD(x1)) = 2 + x1   
POL(false0) = 1   
POL(if_mod(x1)) = 2 + 2·x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(mod(x1)) = 2 + 2·x1   
POL(pred0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 2   

(33) Obligation:

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

MOD(s0(x)) → IF_MOD(le, s0(x))
IF_MOD(true0, s0(x)) → MOD(minus(x))

The TRS R consists of the following rules:

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0
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.


MOD(s0(x)) → IF_MOD(le, s0(x))
IF_MOD(true0, s0(x)) → MOD(minus(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MOD(x1)  =  MOD(x1)
s0(x1)  =  s0(x1)
IF_MOD(x1, x2)  =  IF_MOD(x1, x2)
le  =  le
true0  =  true0
minus(x1)  =  x1
rand0(x1)  =  rand0
pred0(x1)  =  x1
mod(x1)  =  mod(x1)
if_mod(x1)  =  if_mod(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[MOD1, s01, IFMOD2, mod1, ifmod1] > [le, true0]
rand0 > [le, true0]

Status:
MOD1: [1]
s01: [1]
IFMOD2: [1,2]
le: multiset
true0: multiset
rand0: []
mod1: multiset
ifmod1: multiset


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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0
minus(x) → x

(35) Obligation:

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

le → le
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
mod(s0(x)) → if_mod(s0(x))
minus(x) → pred0(minus(x))
if_mod(s0(x)) → mod(minus(x))
le → true0
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