YES Termination proof of rtL-wl1nz.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPCleverAfsProof (⇒, 51 ms)
↳6 QDP
↳7 MRRProof (⇔, 0 ms)
↳8 QDP
↳9 MRRProof (⇔, 5 ms)
↳10 QDP
↳11 MRRProof (⇔, 0 ms)
↳12 QDP
↳13 PisEmptyProof (⇔, 0 ms)
↳14 YES

(0) Obligation:

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

top(ok(new(x))) → top(check(x))
top(ok(old(x))) → top(check(x))

The relative TRS consists of the following S rules:

check(old(x)) → ok(old(x))
bot → new(bot)
check(old(x)) → old(check(x))
check(new(x)) → new(check(x))
old(ok(x)) → ok(old(x))
new(ok(x)) → ok(new(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:

top(ok(new(x))) → TOP(check(x))
top(ok(new(x))) → top(CHECK(x))
top(ok(old(x))) → TOP(check(x))
top(ok(old(x))) → top(CHECK(x))

and relative ADPs:

check(old(x)) → ok(OLD(x))
bot → NEW(BOT)
check(old(x)) → OLD(CHECK(x))
check(new(x)) → NEW(CHECK(x))
old(ok(x)) → ok(OLD(x))
new(ok(x)) → ok(NEW(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:

top(ok(old(x))) → TOP(check(x))
top(ok(new(x))) → top(check(x))
top(ok(old(x))) → top(check(x))
top(ok(new(x))) → TOP(check(x))

and relative ADPs:

check(old(x)) → ok(old(x))
bot → new(bot)
check(old(x)) → old(check(x))
check(new(x)) → new(check(x))
old(ok(x)) → ok(old(x))
new(ok(x)) → ok(new(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:old_1 = new_1 = top_1 = check_1 = bot = ok_1 = TOP_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.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
new(x1)  =  new(x1)
check(x1)  =  x1
old(x1)  =  x1
top(x1)  =  x1

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

[TOP1, new1]

Status:
TOP1: [1]
new1: [1]

(6) Obligation:

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

TOP0(ok0(new0(x))) → TOP0(check0(x))
TOP0(ok0(old0(x))) → TOP0(check0(x))

The TRS R consists of the following rules:

check0(old0(x)) → ok0(old0(x))
bot0 → new0(bot0)
check0(old0(x)) → old0(check0(x))
check0(new0(x)) → new0(check0(x))
old0(ok0(x)) → ok0(old0(x))
top0(ok0(new0(x))) → top0(check0(x))
top0(ok0(old0(x))) → top0(check0(x))
new0(ok0(x)) → ok0(new0(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:

TOP0(ok0(old0(x))) → TOP0(check0(x))

Strictly oriented rules of the TRS R:

top0(ok0(old0(x))) → top0(check0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(TOP0(x1)) = x1   
POL(bot0) = 1   
POL(check0(x1)) = 1 + 2·x1   
POL(new0(x1)) = x1   
POL(ok0(x1)) = 1 + 2·x1   
POL(old0(x1)) = 1 + 2·x1   
POL(top0(x1)) = 2·x1   

(8) Obligation:

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

TOP0(ok0(new0(x))) → TOP0(check0(x))

The TRS R consists of the following rules:

check0(old0(x)) → ok0(old0(x))
bot0 → new0(bot0)
check0(old0(x)) → old0(check0(x))
check0(new0(x)) → new0(check0(x))
old0(ok0(x)) → ok0(old0(x))
top0(ok0(new0(x))) → top0(check0(x))
new0(ok0(x)) → ok0(new0(x))

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 rules of the TRS R:

check0(old0(x)) → ok0(old0(x))
check0(old0(x)) → old0(check0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(TOP0(x1)) = 2·x1   
POL(bot0) = 0   
POL(check0(x1)) = 2·x1   
POL(new0(x1)) = 2·x1   
POL(ok0(x1)) = x1   
POL(old0(x1)) = 2 + x1   
POL(top0(x1)) = 2·x1   

(10) Obligation:

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

TOP0(ok0(new0(x))) → TOP0(check0(x))

The TRS R consists of the following rules:

bot0 → new0(bot0)
check0(new0(x)) → new0(check0(x))
old0(ok0(x)) → ok0(old0(x))
top0(ok0(new0(x))) → top0(check0(x))
new0(ok0(x)) → ok0(new0(x))

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

(11) 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:

TOP0(ok0(new0(x))) → TOP0(check0(x))

Strictly oriented rules of the TRS R:

old0(ok0(x)) → ok0(old0(x))
top0(ok0(new0(x))) → top0(check0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(TOP0(x1)) = x1   
POL(bot0) = 2   
POL(check0(x1)) = x1   
POL(new0(x1)) = x1   
POL(ok0(x1)) = 2 + x1   
POL(old0(x1)) = 1 + 2·x1   
POL(top0(x1)) = x1   

(12) Obligation:

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

bot0 → new0(bot0)
check0(new0(x)) → new0(check0(x))
new0(ok0(x)) → ok0(new0(x))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) YES