YES Termination proof of INVY_15_3.56_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 (⇒, 29 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 39 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 RelADPP
        ↳17 RelADPCleverAfsProof (⇒, 31 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 38 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 RelADPP
        ↳28 RelADPCleverAfsProof (⇒, 41 ms)
        ↳29 QDP
        ↳30 MRRProof (⇔, 35 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 PisEmptyProof (⇔, 0 ms)
        ↳35 YES

(0) Obligation:

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

g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → 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:

g(c(x, s(y))) → G(c(s(x), y))
f(c(s(x), y)) → F(c(x, s(y)))
f(f(x)) → F(d(f(x)))
f(f(x)) → f(d(F(x)))
f(x) → 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:
3 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 3 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

g(c(x, s(y))) → G(c(s(x), y))

and relative ADPs:

f(x) → x
f(f(x)) → f(d(f(x)))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
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_2 = 0 G_1 = d_1 = f_1 = rand_1 = g_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.
G(x1)  =  x1
c(x1, x2)  =  c(x2)
s(x1)  =  s(x1)

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

[c1, s1]

Status:
c1: [1]
s1: multiset

(7) Obligation:

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

G0(c(s0(y))) → G0(c(y))

The TRS R consists of the following rules:

g0(c(s0(y))) → g0(c(y))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(y)) → f0(c(s0(y)))
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(G0(x1)) = x1   
POL(c(x1)) = x1   
POL(d0(x1)) = x1   
POL(f0(x1)) = 2·x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(9) Obligation:

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

G0(c(s0(y))) → G0(c(y))

The TRS R consists of the following rules:

g0(c(s0(y))) → g0(c(y))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(y)) → f0(c(s0(y)))

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

(10) 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:

f0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(G0(x1)) = x1   
POL(c(x1)) = x1   
POL(d0(x1)) = x1   
POL(f0(x1)) = 1 + 2·x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(11) Obligation:

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

G0(c(s0(y))) → G0(c(y))

The TRS R consists of the following rules:

g0(c(s0(y))) → g0(c(y))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(y)) → f0(c(s0(y)))

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

(12) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


G0(c(s0(y))) → G0(c(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G0(x1)  =  G0(x1)
c(x1)  =  c(x1)
s0(x1)  =  s0(x1)
g0(x1)  =  x1
f0(x1)  =  f0
d0(x1)  =  d0(x1)
rand0(x1)  =  rand0

Recursive path order with status [RPO].
Quasi-Precedence:
[G01, c1, s01, f0] > d01

Status:
G01: multiset
c1: multiset
s01: multiset
f0: []
d01: multiset
rand0: multiset


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

g0(c(s0(y))) → g0(c(y))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(y)) → f0(c(s0(y)))

(13) Obligation:

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

g0(c(s0(y))) → g0(c(y))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(y)) → f0(c(s0(y)))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

(16) Obligation:

Relative ADP Problem with
absolute ADPs:

f(c(s(x), y)) → F(c(x, s(y)))

and relative ADPs:

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

(17) 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_2 = 1 d_1 = f_1 = F_1 = rand_1 = g_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)  =  x1
c(x1, x2)  =  c(x1)
s(x1)  =  s(x1)

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

[c1, s1]

Status:
c1: [1]
s1: multiset

(18) Obligation:

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

F0(c(s0(x))) → F0(c(x))

The TRS R consists of the following rules:

g0(c(x)) → g0(c(s0(x)))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(s0(x))) → f0(c(x))
rand0(x) → x

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 rules of the TRS R:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1)) = x1   
POL(c(x1)) = 2·x1   
POL(d0(x1)) = x1   
POL(f0(x1)) = 2·x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(20) Obligation:

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

F0(c(s0(x))) → F0(c(x))

The TRS R consists of the following rules:

g0(c(x)) → g0(c(s0(x)))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(s0(x))) → f0(c(x))

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

(21) 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:

f0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1)) = x1   
POL(c(x1)) = 2·x1   
POL(d0(x1)) = x1   
POL(f0(x1)) = 1 + x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(22) Obligation:

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

F0(c(s0(x))) → F0(c(x))

The TRS R consists of the following rules:

g0(c(x)) → g0(c(s0(x)))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(s0(x))) → f0(c(x))

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

(23) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F0(c(s0(x))) → F0(c(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F0(x1)  =  x1
c(x1)  =  c(x1)
s0(x1)  =  s0(x1)
g0(x1)  =  g0
f0(x1)  =  f0(x1)
d0(x1)  =  x1
rand0(x1)  =  rand0

Recursive path order with status [RPO].
Quasi-Precedence:
rand0 > [s01, f01] > [c1, g0]

Status:
c1: multiset
s01: [1]
g0: multiset
f01: [1]
rand0: multiset


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

g0(c(x)) → g0(c(s0(x)))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(s0(x))) → f0(c(x))

(24) Obligation:

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

g0(c(x)) → g0(c(s0(x)))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c(s0(x))) → f0(c(x))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) YES

(27) Obligation:

Relative ADP Problem with
absolute ADPs:

f(f(x)) → f(d(F(x)))

and relative ADPs:

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

(28) 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_2 = d_1 = f_1 = F_1 = rand_1 = g_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)  =  x1
f(x1)  =  f(x1)

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

trivial

Status:
f1: multiset

(29) Obligation:

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

F0(f0(x)) → F0(x)

The TRS R consists of the following rules:

g0(c0(x, s0(y))) → g0(c0(s0(x), y))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c0(s0(x), y)) → f0(c0(x, s0(y)))
rand0(x) → x

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

(30) 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)) = x1   
POL(c0(x1, x2)) = 2·x1 + x2   
POL(d0(x1)) = x1   
POL(f0(x1)) = 2·x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(31) Obligation:

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

F0(f0(x)) → F0(x)

The TRS R consists of the following rules:

g0(c0(x, s0(y))) → g0(c0(s0(x), y))
f0(x) → x
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c0(s0(x), y)) → f0(c0(x, s0(y)))

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

(32) 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(f0(x)) → F0(x)

Strictly oriented rules of the TRS R:

f0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1)) = x1   
POL(c0(x1, x2)) = 2·x1 + 2·x2   
POL(d0(x1)) = x1   
POL(f0(x1)) = 1 + 2·x1   
POL(g0(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(33) Obligation:

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

g0(c0(x, s0(y))) → g0(c0(s0(x), y))
f0(f0(x)) → f0(d0(f0(x)))
rand0(x) → rand0(s0(x))
f0(c0(s0(x), y)) → f0(c0(x, s0(y)))

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) YES