Termination proof of Relative_05

Termination of the given RelTRS could be proven:

0 RelTRS

↳1 RelTRStoRelADPProof (⇔, 0 ms)

↳2 RelADPP

↳3 RelADPDepGraphProof (⇔, 0 ms)

↳4 RelADPP

↳5 RelADPCleverAfsProof (⇒, 48 ms)

↳6 QDP

↳7 MRRProof (⇔, 0 ms)

↳8 QDP

↳9 PisEmptyProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

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

t(f(x), g(y), f(z)) → t(z, g(x), g(y))
t(g(x), g(y), f(z)) → t(f(y), f(z), x)

The relative TRS consists of the following S rules:

g(g(x)) → f(f(x))
f(f(x)) → g(g(x))
g(f(x)) → f(g(x))
f(g(x)) → g(f(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:

t(f(x), g(y), f(z)) → T(z, g(x), g(y))
t(f(x), g(y), f(z)) → t(z, G(x), g(y))
t(f(x), g(y), f(z)) → t(z, g(x), G(y))
t(g(x), g(y), f(z)) → T(f(y), f(z), x)
t(g(x), g(y), f(z)) → t(F(y), f(z), x)
t(g(x), g(y), f(z)) → t(f(y), F(z), x)

and relative ADPs:

g(g(x)) → F(F(x))
f(f(x)) → G(G(x))
g(f(x)) → F(G(x))
f(g(x)) → G(F(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:

t(g(x), g(y), f(z)) → T(f(y), f(z), x)
t(f(x), g(y), f(z)) → T(z, g(x), g(y))

and relative ADPs:

t(f(x), g(y), f(z)) → t(z, g(x), g(y))
g(g(x)) → f(f(x))
f(f(x)) → g(g(x))
g(f(x)) → f(g(x))
t(g(x), g(y), f(z)) → t(f(y), f(z), x)
f(g(x)) → g(f(x))

(5) 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:t_3 = f_1 = T_3 = g_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Recursive path order with status [RPO].
Quasi-Precedence:

[f₁, g₁] > T₃

Status:

T₃: multiset
f₁: multiset
g₁: multiset

(6) Obligation:

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

T0(f0(x), g0(y), f0(z)) → T0(z, g0(x), g0(y))
T0(g0(x), g0(y), f0(z)) → T0(f0(y), f0(z), x)

The TRS R consists of the following rules:

t0(f0(x), g0(y), f0(z)) → t0(z, g0(x), g0(y))
g0(g0(x)) → f0(f0(x))
f0(f0(x)) → g0(g0(x))
g0(f0(x)) → f0(g0(x))
t0(g0(x), g0(y), f0(z)) → t0(f0(y), f0(z), x)
f0(g0(x)) → g0(f0(x))

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

(7) 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:

T0(f0(x), g0(y), f0(z)) → T0(z, g0(x), g0(y))
T0(g0(x), g0(y), f0(z)) → T0(f0(y), f0(z), x)

Strictly oriented rules of the TRS R:

t0(f0(x), g0(y), f0(z)) → t0(z, g0(x), g0(y))
t0(g0(x), g0(y), f0(z)) → t0(f0(y), f0(z), x)

Used ordering: Polynomial interpretation [POLO]:

POL(T0(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + 2·x₃
POL(f0(x₁)) = 1 + x₁
POL(g0(x₁)) = 1 + x₁
POL(t0(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + 2·x₃

(8) Obligation:

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

g0(g0(x)) → f0(f0(x))
f0(f0(x)) → g0(g0(x))
g0(f0(x)) → f0(g0(x))
f0(g0(x)) → g0(f0(x))