YES Termination proof of AG_#3.54_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 (⇒, 46 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 3 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 RelADPP
        ↳15 RelADPCleverAfsProof (⇒, 46 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES

(0) Obligation:

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

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, 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:

f(g(x)) → g(F(f(x)))
f(g(x)) → g(f(F(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → F'(y, x, 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:
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(g(x)) → g(f(f(x)))
f'(s(x), y, y) → F'(y, x, s(x))
f(h(x)) → h(g(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 = F'_3 = 2 f_1 = h_1 = rand_1 = g_1 = f'_3 = 0, 1, 2
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, x3)  =  F'(x1, x2)
s(x1)  =  s(x1)
f(x1)  =  x1
g(x1)  =  x1
f'(x1, x2, x3)  =  f'
h(x1)  =  h(x1)

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

f' > [F'2, s1]

Status:
F'2: multiset
s1: multiset
f': multiset
h1: multiset

(7) Obligation:

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

F'(s0(x), y) → F'(y, x)

The TRS R consists of the following rules:

f0(g0(x)) → g0(f0(f0(x)))
f' → f'
f0(h0(x)) → h0(g0(x))
rand0(x) → rand0(s0(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, x2)) = x1 + x2   
POL(f') = 0   
POL(f0(x1)) = x1   
POL(g0(x1)) = x1   
POL(h0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(9) Obligation:

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

F'(s0(x), y) → F'(y, x)

The TRS R consists of the following rules:

f0(g0(x)) → g0(f0(f0(x)))
f' → f'
f0(h0(x)) → h0(g0(x))
rand0(x) → rand0(s0(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), y) → F'(y, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F'(x1, x2)  =  F'(x1, x2)
s0(x1)  =  s0(x1)
f0(x1)  =  x1
g0(x1)  =  x1
f'  =  f'
h0(x1)  =  x1
rand0(x1)  =  rand0

Recursive path order with status [RPO].
Quasi-Precedence:
rand0 > [F'2, s01]

Status:
F'2: multiset
s01: multiset
f': multiset
rand0: multiset


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

f0(g0(x)) → g0(f0(f0(x)))
f' → f'
f0(h0(x)) → h0(g0(x))
rand0(x) → rand0(s0(x))

(11) Obligation:

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

f0(g0(x)) → g0(f0(f0(x)))
f' → f'
f0(h0(x)) → h0(g0(x))
rand0(x) → rand0(s0(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(g(x)) → g(F(f(x)))
f'(s(x), y, y) → f'(y, x, s(x))
f(h(x)) → h(g(x))
f(g(x)) → g(f(F(x)))

and relative ADPs:

rand(x) → rand(s(x))
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 f_1 = h_1 = 0 F_1 = rand_1 = g_1 = f'_3 = 0, 1, 2
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)  =  F(x1)
g(x1)  =  g(x1)
f(x1)  =  x1
f'(x1, x2, x3)  =  f'
s(x1)  =  s
h(x1)  =  h

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

f' > s
h > [F1, g1]

Status:
F1: [1]
g1: multiset
f': multiset
s: multiset
h: multiset

(16) Obligation:

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

F0(g0(x)) → F0(x)
F0(g0(x)) → F0(f0(x))

The TRS R consists of the following rules:

f0(g0(x)) → g0(f0(f0(x)))
f' → f'
f0(h) → h
rand0(x) → rand0(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 dependency pairs:

F0(g0(x)) → F0(x)
F0(g0(x)) → F0(f0(x))

Strictly oriented rules of the TRS R:

f0(g0(x)) → g0(f0(f0(x)))
f0(h) → h
rand0(x) → x

Used ordering: Knuth-Bendix order [KBO] with precedence:
f01 > F01 > h > f' > s > rand01 > g01

and weight map:

f'=1
h=1
s=1
f0_1=0
g0_1=1
rand0_1=1
F0_1=1

The variable weight is 1

(18) Obligation:

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

f' → f'
rand0(x) → rand0(s)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES