Termination proof of AG_#3.6a

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 (⇒, 128 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 37 ms)

        ↳9 QDP

        ↳10 QDPOrderProof (⇔, 14 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 129 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 49 ms)

        ↳18 QDP

        ↳19 QDPOrderProof (⇔, 19 ms)

        ↳20 QDP

        ↳21 PisEmptyProof (⇔, 0 ms)

        ↳22 YES

    ↳23 RelADPP

        ↳24 RelADPCleverAfsProof (⇒, 127 ms)

        ↳25 QDP

        ↳26 MRRProof (⇔, 35 ms)

        ↳27 QDP

        ↳28 MRRProof (⇔, 0 ms)

        ↳29 QDP

        ↳30 QDPOrderProof (⇔, 0 ms)

        ↳31 QDP

        ↳32 PisEmptyProof (⇔, 0 ms)

        ↳33 YES

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → LE(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
gcd(s(x), s(y)) → if_gcd(LE(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
if_gcd(true, s(x), s(y)) → gcd(MINUS(x, y), s(y))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))
if_gcd(false, s(x), s(y)) → gcd(MINUS(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:
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:

gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(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 = MINUS_2 = 0 true = gcd_2 = 0 = le_2 = if_gcd_3 = 0 minus_2 = 1 rand_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.
MINUS(x1, x2) = MINUS(x2)
s(x1) = s(x1)
gcd(x1, x2) = gcd(x1, x2)
0 = 0
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
le(x1, x2) = le(x1, x2)
false = false
true = true
minus(x1, x2) = minus(x1)

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

0 > [s₁, minus₁] > MINUS₁ > [gcd₂, if_gcd₂, le₂]
0 > [s₁, minus₁] > false > [gcd₂, if_gcd₂, le₂]
0 > true > [gcd₂, if_gcd₂, le₂]

Status:

MINUS₁: [1]
s₁: multiset
gcd₂: multiset
0: multiset
if_gcd₂: multiset
le₂: multiset
false: multiset
true: multiset
minus₁: multiset

(7) Obligation:

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

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
le0(s0(x), 00) → false0
gcd0(s0(x), 00) → s0(x)
le0(00, y) → true0
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), 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:

gcd0(00, y) → y
le0(s0(x), 00) → false0
gcd0(s0(x), 00) → s0(x)
le0(00, y) → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1
POL(MINUS(x₁)) = x₁
POL(false0) = 0
POL(gcd0(x₁, x₂)) = 2·x₁ + 2·x₂
POL(if_gcd(x₁, x₂)) = 2·x₁ + 2·x₂
POL(le0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
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:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), 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.

MINUS(s0(y)) → MINUS(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( gcd0(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( s0(x₁) ) = 2x₁ + 1

POL( if_gcd(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( le0(x₁, x₂) ) = 0

POL( minus(x₁) ) = x₁

POL( rand0(x₁) ) = max{0, -2}

POL( MINUS(x₁) ) = x₁

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

(11) Obligation:

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), 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:

gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
le(s(x), s(y)) → LE(x, y)

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 = true = gcd_2 = LE_2 = 0 0 = le_2 = if_gcd_3 = 0 minus_2 = 1 rand_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.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)
gcd(x1, x2) = gcd(x1, x2)
0 = 0
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
le(x1, x2) = le(x1, x2)
false = false
true = true
minus(x1, x2) = minus(x1)

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

0 > [s₁, minus₁] > LE₁ > [gcd₂, if_gcd₂, le₂]
0 > [s₁, minus₁] > false > [gcd₂, if_gcd₂, le₂]
0 > true > [gcd₂, if_gcd₂, le₂]

Status:

LE₁: [1]
s₁: multiset
gcd₂: multiset
0: multiset
if_gcd₂: multiset
le₂: multiset
false: multiset
true: multiset
minus₁: multiset

(16) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
le0(s0(x), 00) → false0
gcd0(s0(x), 00) → s0(x)
le0(00, y) → true0
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
rand0(x) → x

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

(17) MRRProof (EQUIVALENT transformation)

gcd0(00, y) → y
le0(s0(x), 00) → false0
gcd0(s0(x), 00) → s0(x)
le0(00, y) → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1
POL(LE(x₁)) = x₁
POL(false0) = 0
POL(gcd0(x₁, x₂)) = 2·x₁ + 2·x₂
POL(if_gcd(x₁, x₂)) = 2·x₁ + 2·x₂
POL(le0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
POL(rand0(x₁)) = 2 + x₁
POL(s0(x₁)) = x₁
POL(true0) = 0

(18) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

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

(19) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

LE(s0(y)) → LE(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1) = x1
s0(x1) = s0(x1)
gcd0(x1, x2) = gcd0(x1, x2)
if_gcd(x1, x2) = if_gcd(x1, x2)
le0(x1, x2) = le0
minus(x1) = minus(x1)
rand0(x1) = rand0

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

rand0 > [s0₁, minus₁] > [gcd0₂, if_gcd₂]

Status:

s0₁: multiset
gcd0₂: multiset
if_gcd₂: multiset
le0: []
minus₁: multiset
rand0: []

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

(20) Obligation:

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le0(s0(x), s0(y)) → le0(x, y)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES

(23) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(x), s(y)) → le(x, y)
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(24) 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 = IF_GCD_3 = 0 gcd_2 = le_2 = 0, 1 0 = if_gcd_3 = 0 minus_2 = 1 rand_1 = false = GCD_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.
IF_GCD(x1, x2, x3) = IF_GCD(x2, x3)
false = false
s(x1) = s(x1)
GCD(x1, x2) = GCD(x1, x2)
minus(x1, x2) = x1
le(x1, x2) = le
true = true
gcd(x1, x2) = gcd(x1, x2)
0 = 0
if_gcd(x1, x2, x3) = if_gcd(x2, x3)

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

le > [IF_GCD₂, false, GCD₂] > [gcd₂, if_gcd₂] > s₁
le > true
0 > [IF_GCD₂, false, GCD₂] > [gcd₂, if_gcd₂] > s₁
0 > true

Status:

IF_GCD₂: multiset
false: multiset
s₁: [1]
GCD₂: multiset
le: multiset
true: multiset
gcd₂: multiset
0: multiset
if_gcd₂: multiset

(25) Obligation:

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

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
le → true0
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
rand0(x) → x

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

(26) MRRProof (EQUIVALENT transformation)

le → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(GCD0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(IF_GCD(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(false0) = 2
POL(gcd0(x₁, x₂)) = 2·x₁ + 2·x₂
POL(if_gcd(x₁, x₂)) = 2·x₁ + 2·x₂
POL(le) = 2
POL(minus(x₁)) = x₁
POL(rand0(x₁)) = 2 + x₁
POL(s0(x₁)) = x₁
POL(true0) = 0

(27) Obligation:

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

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

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

(28) MRRProof (EQUIVALENT transformation)

gcd0(00, y) → y
le → false0
gcd0(s0(x), 00) → s0(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1
POL(GCD0(x₁, x₂)) = 2 + x₁ + x₂
POL(IF_GCD(x₁, x₂)) = 2 + x₁ + x₂
POL(false0) = 1
POL(gcd0(x₁, x₂)) = x₁ + x₂
POL(if_gcd(x₁, x₂)) = x₁ + x₂
POL(le) = 2
POL(minus(x₁)) = x₁
POL(rand0(x₁)) = x₁
POL(s0(x₁)) = x₁

(29) Obligation:

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

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → le
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

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

(30) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( gcd0(x₁, x₂) ) = 2x₁ + 2x₂

POL( s0(x₁) ) = 2x₁ + 2

POL( if_gcd(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 2}

POL( le ) = 0

POL( minus(x₁) ) = 2x₁

POL( rand0(x₁) ) = 1

POL( IF_GCD(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( GCD0(x₁, x₂) ) = 2x₁ + 2x₂

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → le
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

(31) Obligation:

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → le
minus(s0(x)) → minus(x)
minus(x) → x
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
rand0(x) → rand0(s0(x))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))

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

(32) 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) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) YES

(23) Obligation:

(24) RelADPCleverAfsProof (SOUND transformation)

(25) Obligation:

(26) MRRProof (EQUIVALENT transformation)

(27) Obligation:

(28) MRRProof (EQUIVALENT transformation)

(29) Obligation:

(30) QDPOrderProof (EQUIVALENT transformation)

(31) Obligation:

(32) PisEmptyProof (EQUIVALENT transformation)

(33) YES