Termination proof of INVY_15_3.48

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 (⇒, 44 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 24 ms)

        ↳9 QDP

        ↳10 QDPOrderProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 51 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(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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(0) → true
f(1) → false
f(s(x)) → F(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(G(x, y))
g(x, c(y)) → G(x, if(f(x), c(g(s(x), y)), c(y)))
g(x, c(y)) → g(x, IF(f(x), c(g(s(x), y)), c(y)))
g(x, c(y)) → g(x, if(F(x), c(g(s(x), y)), c(y)))
g(x, c(y)) → g(x, if(f(x), c(G(s(x), y)), c(y)))

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(s(x)) → F(x)

and relative ADPs:

if(false, s(x), s(y)) → s(y)
if(true, s(x), s(y)) → s(x)
g(x, c(y)) → c(g(x, y))
f(1) → false
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))
f(0) → true
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 = true = c_1 = f_1 = if_3 = 0 = 1 = F_1 = rand_1 = g_2 = 1 false =
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
s(x1) = s(x1)
g(x1, x2) = g(x1)
c(x1) = x1
if(x1, x2, x3) = if(x1, x2, x3)
f(x1) = f(x1)
false = false
true = true
1 = 1
0 = 0

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

g₁ > s₁
g₁ > if₃
[f₁, true] > s₁
[f₁, true] > false
1 > false

Status:

s₁: multiset
g₁: [1]
if₃: multiset
f₁: multiset
false: multiset
true: multiset
1: multiset
0: multiset

(7) Obligation:

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

F0(s0(x)) → F0(x)

The TRS R consists of the following rules:

if0(false0, s0(x), s0(y)) → s0(y)
if0(true0, s0(x), s0(y)) → s0(x)
g(x) → c0(g(x))
f0(10) → false0
g(x) → g(x)
f0(00) → true0
f0(s0(x)) → f0(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:

if0(false0, s0(x), s0(y)) → s0(y)
if0(true0, s0(x), s0(y)) → s0(x)
f0(10) → false0
f0(00) → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(10) = 0
POL(F0(x₁)) = x₁
POL(c0(x₁)) = x₁
POL(f0(x₁)) = 1 + x₁
POL(false0) = 0
POL(g(x₁)) = x₁
POL(if0(x₁, x₂, x₃)) = 2 + x₁ + x₂ + x₃
POL(rand0(x₁)) = 2 + x₁
POL(s0(x₁)) = x₁
POL(true0) = 0

(9) Obligation:

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

F0(s0(x)) → F0(x)

The TRS R consists of the following rules:

g(x) → c0(g(x))
g(x) → g(x)
f0(s0(x)) → f0(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.

F0(s0(x)) → F0(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F0(x1) = x1
s0(x1) = s0(x1)
g(x1) = g
c0(x1) = c0
f0(x1) = f0
rand0(x1) = rand0

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

[s0₁, f0] > rand0
[g, c0] > rand0

Status:

s0₁: multiset
g: []
c0: []
f0: multiset
rand0: multiset

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

g(x) → c0(g(x))
g(x) → g(x)
f0(s0(x)) → f0(x)
rand0(x) → rand0(s0(x))

(11) Obligation:

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

g(x) → c0(g(x))
g(x) → g(x)
f0(s0(x)) → f0(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:

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

and relative ADPs:

if(false, s(x), s(y)) → s(y)
if(true, s(x), s(y)) → s(x)
f(1) → false
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))
f(0) → true
f(s(x)) → f(x)
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 true = c_1 = G_2 = 0 if_3 = 0, 1, 2 f_1 = 0 0 = 1 = rand_1 = g_2 = 0 false =
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, x2) = G(x2)
c(x1) = c(x1)
s(x1) = s
g(x1, x2) = g(x2)
if(x1, x2, x3) = if
f(x1) = f
false = false
true = true
1 = 1
0 = 0

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

g₁ > [G₁, c₁, s, if]
f > false > [G₁, c₁, s, if]
f > true > [G₁, c₁, s, if]
0 > true > [G₁, c₁, s, if]

Status:

G₁: multiset
c₁: multiset
s: multiset
g₁: multiset
if: multiset
f: []
false: multiset
true: multiset
1: multiset
0: multiset

(16) Obligation:

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

G(c0(y)) → G(y)

The TRS R consists of the following rules:

if → s
g(c0(y)) → c0(g(y))
f → false0
g(c0(y)) → g(if)
f → true0
f → f
rand0(x) → rand0(s)
rand0(x) → x

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

(17) MRRProof (EQUIVALENT transformation)

G(c0(y)) → G(y)

Strictly oriented rules of the TRS R:

if → s
g(c0(y)) → c0(g(y))
f → false0
g(c0(y)) → g(if)
f → true0
rand0(x) → x

Used ordering: Knuth-Bendix order [KBO] with precedence:

false0 > f > g₁ > c0₁ > G₁ > rand0₁ > if > true0 > s

and weight map:

if=2
s=1
f=2
false0=1
true0=2
g_1=2
c0_1=1
rand0_1=1
G_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.

(0) Obligation:

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

(2) Obligation:

(3) RelADPDepGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) RelADPCleverAfsProof (SOUND transformation)

(7) Obligation:

(8) MRRProof (EQUIVALENT transformation)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) RelADPCleverAfsProof (SOUND transformation)

(16) Obligation:

(17) MRRProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) YES