Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
REVERSE(cons(x, xs)) → LAST(cons(x, xs))
REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))
REVERSE(cons(x, xs)) → DEL(last(cons(x, xs)), cons(x, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

(11) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EQ(s(x), s(y)) → EQ(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
reverse(nil)
reverse(cons(x0, x1))

(18) Obligation:

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF(false, x, y, xs) → DEL(x, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

(20) TRUE

(21) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

R is empty.
The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

(25) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
  The graph contains the following edges 1 > 1

(27) TRUE

(28) Obligation:

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

reverse(nil)
reverse(cons(x0, x1))

(32) Obligation:

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(33) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 2   
POL(REVERSE(x₁)) = x₁   
POL(cons(x₁, x₂)) = 1 + x₁ + 2·x₂   
POL(del(x₁, x₂)) = x₂   
POL(eq(x₁, x₂)) = 2·x₂   
POL(false) = 1   
POL(if(x₁, x₂, x₃, x₄)) = 1 + x₃ + 2·x₄   
POL(last(x₁)) = 3 + 2·x₁   
POL(nil) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 2   

At least one of these decreasing rules is always used after the deleted DP:
if(true, x595, y445, xs285) → xs285


The following formula is valid:
z0:sort[a24].(¬(z0 =nil)→del'(last(z0 ), z0 )=true)


The transformed set:
del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

(34) Complex Obligation (AND)

(35) Obligation:

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

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

Q is empty.

(39) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(40) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(41) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(42) Obligation:

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

DEL'(x16, cons(y11, xs7)) → IF'(eq(x16, y11), x16, y11, xs7)
DEL'(x16, cons(y11, xs7)) → EQ(x16, y11)
IF'(false, x68, y51, xs33) → DEL'(x68, xs33)
LAST(cons(x7, cons(y4, xs2))) → LAST(cons(y4, xs2))
DEL(x16, cons(y11, xs7)) → IF(eq(x16, y11), x16, y11, xs7)
DEL(x16, cons(y11, xs7)) → EQ(x16, y11)
EQ(s(x50), s(y37)) → EQ(x50, y37)
IF(false, x68, y51, xs33) → DEL(x68, xs33)
EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x0, x2)
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x1, x3)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(43) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 3 less nodes.

(44) Complex Obligation (AND)

(45) Obligation:

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

EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x1, x3)
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x0, x2)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(46) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(47) Obligation:

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

EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x1, x3)
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x0, x2)

R is empty.
The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(48) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(49) Obligation:

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

EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x1, x3)
EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x0, x2)

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

(50) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x1, x3)
  The graph contains the following edges 1 > 1, 2 > 2

• EQUAL_SORT[A24](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A24](x0, x2)
  The graph contains the following edges 1 > 1, 2 > 2

(51) TRUE

(52) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(53) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(54) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

R is empty.
The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(55) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(56) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

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

(57) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
  The graph contains the following edges 1 > 1, 2 > 2

(58) TRUE

(59) Obligation:

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

EQ(s(x50), s(y37)) → EQ(x50, y37)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(60) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(61) Obligation:

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

EQ(s(x50), s(y37)) → EQ(x50, y37)

R is empty.
The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(62) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(63) Obligation:

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

EQ(s(x50), s(y37)) → EQ(x50, y37)

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

(64) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EQ(s(x50), s(y37)) → EQ(x50, y37)
  The graph contains the following edges 1 > 1, 2 > 2

(65) TRUE

(66) Obligation:

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

IF(false, x68, y51, xs33) → DEL(x68, xs33)
DEL(x16, cons(y11, xs7)) → IF(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(67) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(68) Obligation:

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

IF(false, x68, y51, xs33) → DEL(x68, xs33)
DEL(x16, cons(y11, xs7)) → IF(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(69) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(70) Obligation:

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

IF(false, x68, y51, xs33) → DEL(x68, xs33)
DEL(x16, cons(y11, xs7)) → IF(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(71) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• DEL(x16, cons(y11, xs7)) → IF(eq(x16, y11), x16, y11, xs7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF(false, x68, y51, xs33) → DEL(x68, xs33)
  The graph contains the following edges 2 >= 1, 4 >= 2

(72) TRUE

(73) Obligation:

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

LAST(cons(x7, cons(y4, xs2))) → LAST(cons(y4, xs2))

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(74) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(75) Obligation:

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

LAST(cons(x7, cons(y4, xs2))) → LAST(cons(y4, xs2))

R is empty.
The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(76) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(77) Obligation:

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

LAST(cons(x7, cons(y4, xs2))) → LAST(cons(y4, xs2))

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

(78) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LAST(cons(x7, cons(y4, xs2))) → LAST(cons(y4, xs2))
  The graph contains the following edges 1 > 1

(79) TRUE

(80) Obligation:

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

IF'(false, x68, y51, xs33) → DEL'(x68, xs33)
DEL'(x16, cons(y11, xs7)) → IF'(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x, nil)) → x
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(81) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(82) Obligation:

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

IF'(false, x68, y51, xs33) → DEL'(x68, xs33)
DEL'(x16, cons(y11, xs7)) → IF'(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

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

(83) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, cons(x1, x2))
if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, nil)
last(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a24](nil, nil)
equal_sort[a24](nil, cons(x0, x1))
equal_sort[a24](cons(x0, x1), nil)
equal_sort[a24](cons(x0, x1), cons(x2, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42])

(84) Obligation:

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

IF'(false, x68, y51, xs33) → DEL'(x68, xs33)
DEL'(x16, cons(y11, xs7)) → IF'(eq(x16, y11), x16, y11, xs7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(85) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• DEL'(x16, cons(y11, xs7)) → IF'(eq(x16, y11), x16, y11, xs7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF'(false, x68, y51, xs33) → DEL'(x68, xs33)
  The graph contains the following edges 2 >= 1, 4 >= 2

(86) TRUE