YES Termination proof of INVY_15_3.47_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 (⇒, 30 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 RelADPP
        ↳15 RelADPCleverAfsProof (⇒, 39 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

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:

f(x, c(y)) → F(x, s(f(y, y)))
f(x, c(y)) → f(x, s(F(y, y)))
f(s(x), y) → F(x, s(c(y)))

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:
2 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 2 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

f(s(x), y) → F(x, s(c(y)))

and relative ADPs:

f(x, c(y)) → f(x, s(f(y, y)))
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 = c_1 = 0 f_2 = 1 F_2 = 1 rand_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.
F(x1, x2)  =  F(x1)
s(x1)  =  s(x1)
c(x1)  =  c
f(x1, x2)  =  f(x1)

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

[F1, s1, c] > f1

Status:
F1: multiset
s1: multiset
c: multiset
f1: [1]

(7) Obligation:

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

F(s0(x)) → F(x)

The TRS R consists of the following rules:

f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(F(x1)) = x1   
POL(f(x1)) = x1   
POL(rand0(x1)) = 1 + x1   
POL(s0(x1)) = x1   

(9) Obligation:

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

F(s0(x)) → F(x)

The TRS R consists of the following rules:

f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F(s0(x)) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
s0(x1)  =  s0(x1)
f(x1)  =  f(x1)
rand0(x1)  =  rand0

Recursive path order with status [RPO].
Quasi-Precedence:
f1 > s01
rand0 > s01

Status:
s01: multiset
f1: multiset
rand0: multiset


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

f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)

(11) Obligation:

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

f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) Obligation:

Relative ADP Problem with
absolute ADPs:

f(x, c(y)) → f(x, s(F(y, y)))

and relative ADPs:

f(x, c(y)) → f(x, s(f(y, y)))
rand(x) → rand(s(x))
f(s(x), y) → f(x, s(c(y)))
rand(x) → x

(15) 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 = 0 c_1 = f_2 = 0 F_2 = rand_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.
F(x1, x2)  =  F(x1, x2)
c(x1)  =  c(x1)
f(x1, x2)  =  x2
s(x1)  =  s

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

F2 > s
c1 > s

Status:
F2: [2,1]
c1: multiset
s: multiset

(16) Obligation:

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

F0(x, c0(y)) → F0(y, y)

The TRS R consists of the following rules:

f(c0(y)) → f(s)
rand0(x) → rand0(s)
f(y) → f(s)
rand0(x) → x

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

(17) 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(F0(x1, x2)) = x1 + 2·x2   
POL(c0(x1)) = 2·x1   
POL(f(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s) = 0   

(18) Obligation:

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

F0(x, c0(y)) → F0(y, y)

The TRS R consists of the following rules:

f(c0(y)) → f(s)
rand0(x) → rand0(s)
f(y) → f(s)

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 dependency pairs:

F0(x, c0(y)) → F0(y, y)

Strictly oriented rules of the TRS R:

f(c0(y)) → f(s)

Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1, x2)) = x1 + 2·x2   
POL(c0(x1)) = 2 + 2·x1   
POL(f(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s) = 0   

(20) Obligation:

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

rand0(x) → rand0(s)
f(y) → f(s)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES