YES Termination proof of ex2.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPDerelatifyingProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 6 ms)
↳8 QDP
↳9 PisEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

g(s(x), y) → g(f(x, y), y)

The relative TRS consists of the following S rules:

f(x, y) → f(x, s(y))
f(x, y) → 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:

g(s(x), y) → G(f(x, y), y)
g(s(x), y) → g(F(x, y), y)

and relative ADPs:

f(x, y) → F(x, s(y))
f(x, y) → x

(3) 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.

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

g(s(x), y) → G(f(x, y), y)
g(s(x), y) → g(f(x, y), y)

and relative ADPs:

f(x, y) → f(x, s(y))
f(x, y) → x

(5) RelADPDerelatifyingProof (EQUIVALENT 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.

(6) Obligation:

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

G(s(x), y) → G(f(x, y), y)

The TRS R consists of the following rules:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


G(s(x), y) → G(f(x, y), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  x1
s(x1)  =  s(x1)
f(x1, x2)  =  f(x1)
g(x1, x2)  =  g(x2)

Recursive path order with status [RPO].
Quasi-Precedence:
s1 > g1 > f1

Status:
s1: multiset
f1: [1]
g1: multiset


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

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

(8) Obligation:

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

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

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) YES