YES Termination proof of INVY_15_ex4.ari

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRSSCleanerProof (⇒, 0 ms)
↳2 RelTRS
↳3 RelTRStoRelADPProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPDepGraphProof (⇔, 0 ms)
↳6 RelADPP
↳7 RelADPCleverAfsProof (⇒, 62 ms)
↳8 QDP
↳9 MRRProof (⇔, 0 ms)
↳10 QDP
↳11 PisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

a → b

The relative TRS consists of the following S rules:

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

(1) RelTRSSCleanerProof (SOUND transformation)

We have moved all duplicating rules from S to R → Duplicating rules:

f(s(x)) → c(x, f(x))

(2) Obligation:

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

a → b
f(s(x)) → c(x, f(x))

The relative TRS consists of the following S rules:

c(x, c(y, z)) → c(y, c(x, z))

(3) RelTRStoRelADPProof (EQUIVALENT transformation)

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

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

a → b
f(s(x)) → C(x, f(x))
f(s(x)) → c(x, F(x))

and relative ADPs:

c(x, c(y, z)) → C(y, C(x, z))

(5) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
1 SCC with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 1 subproblem.

(6) Obligation:

Relative ADP Problem with
absolute ADPs:

f(s(x)) → c(x, F(x))

and relative ADPs:

a → b
c(x, c(y, z)) → c(y, c(x, z))
f(s(x)) → c(x, f(x))

(7) 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 = a = b = c_2 = 0 y = f_1 = 0 F_1 = z =
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)  =  x1
s(x1)  =  s(x1)
f(x1)  =  f
c(x1, x2)  =  x2
y  =  y
z  =  z

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

[y, z]

Status:
s1: multiset
f: multiset
y: multiset
z: multiset

(8) Obligation:

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

F0(s0(x)) → F0(x)

The TRS R consists of the following rules:

a0 → b0
c(c(z0)) → c(c(z0))
f → c(f)

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

(9) 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(s0(x)) → F0(x)

Strictly oriented rules of the TRS R:

a0 → b0

Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1)) = x1   
POL(a0) = 2   
POL(b0) = 1   
POL(c(x1)) = x1   
POL(f) = 0   
POL(s0(x1)) = 1 + x1   
POL(z0) = 0   

(10) Obligation:

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

c(c(z0)) → c(c(z0))
f → c(f)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES