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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(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:

LE(s(x), s(y)) → LE(x, y)
EQ(s(x), s(y)) → EQ(x, y)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)
IF2(false, x, y, xs) → DEL(x, xs)
MINSORT(cons(x, y)) → MIN(x, y)
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))
MINSORT(cons(x, y)) → DEL(min(x, y), cons(x, y))
MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
MIN(x, cons(y, z)) → LE(x, y)
DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
DEL(x, cons(y, z)) → EQ(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(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:

DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
IF2(false, x, y, xs) → DEL(x, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
IF2(false, x, y, xs) → DEL(x, 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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(18) Obligation:

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

DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
IF2(false, x, y, xs) → DEL(x, 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:

• IF2(false, x, y, xs) → DEL(x, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

• DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

(20) TRUE

(21) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(25) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

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:

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

(27) TRUE

(28) Obligation:

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

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(32) Obligation:

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

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(33) 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:

• MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF1(true, x, y, xs) → MIN(x, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

• IF1(false, x, y, xs) → MIN(y, xs)
  The graph contains the following edges 3 >= 1, 4 >= 2

(34) TRUE

(35) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(36) 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.

(37) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
if1(true, x, y, xs) → min(x, xs)
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(38) 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].

minsort(nil)
minsort(cons(x0, x1))

(39) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
if1(true, x, y, xs) → min(x, xs)
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(40) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(MINSORT(x₁)) = x₁   
POL(cons(x₁, x₂)) = 1 + x₁ + x₂   
POL(del(x₁, x₂)) = x₂   
POL(eq(x₁, x₂)) = x₂   
POL(false) = 0   
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₂ + x₃ + x₄   
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄   
POL(le(x₁, x₂)) = 1   
POL(min(x₁, x₂)) = 1 + x₁ + x₂   
POL(nil) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

At least one of these decreasing rules is always used after the deleted DP:
if2(true, x1127, y937, xs357) → xs357


The following formula is valid:
x:sort[a0],y:sort[a32].del'(min(x , y ), cons(x , y ))=true


The transformed set:
del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

(41) Complex Obligation (AND)

(42) Obligation:

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

min(x, nil) → x
if1(true, x, y, xs) → min(x, xs)
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE

(45) Obligation:

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

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

Q is empty.

(46) Overlay + Local Confluence (EQUIVALENT transformation)

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

(47) Obligation:

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

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(48) DependencyPairsProof (EQUIVALENT transformation)

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

(49) Obligation:

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

DEL'(x49, cons(y40, z7)) → IF2'(eq(x49, y40), x49, y40, z7)
DEL'(x49, cons(y40, z7)) → EQ(x49, y40)
IF2'(false, x125, y104, xs40) → DEL'(x125, xs40)
IF1(true, x10, y7, xs1) → MIN(x10, xs1)
MIN(x23, cons(y18, z2)) → IF1(le(x23, y18), x23, y18, z2)
MIN(x23, cons(y18, z2)) → LE(x23, y18)
IF1(false, x36, y29, xs10) → MIN(y29, xs10)
DEL(x49, cons(y40, z7)) → IF2(eq(x49, y40), x49, y40, z7)
DEL(x49, cons(y40, z7)) → EQ(x49, y40)
EQ(s(x99), s(y82)) → EQ(x99, y82)
IF2(false, x125, y104, xs40) → DEL(x125, xs40)
LE(s(x176), s(y146)) → LE(x176, y146)
EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x0, x2)
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x1, x3)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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

(50) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 4 less nodes.

(51) Complex Obligation (AND)

(52) Obligation:

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

EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x1, x3)
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x0, x2)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x1, x3)
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x0, x2)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(56) Obligation:

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

EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x1, x3)
EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x0, x2)

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[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x1, x3)
  The graph contains the following edges 1 > 1, 2 > 2

• EQUAL_SORT[A32](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A32](x0, x2)
  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:

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

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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:

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(63) 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.

(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:

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

LE(s(x176), s(y146)) → LE(x176, y146)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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:

LE(s(x176), s(y146)) → LE(x176, y146)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(70) Obligation:

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

LE(s(x176), s(y146)) → LE(x176, y146)

R is empty.
Q is empty.
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:

• LE(s(x176), s(y146)) → LE(x176, y146)
  The graph contains the following edges 1 > 1, 2 > 2

(72) TRUE

(73) Obligation:

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

EQ(s(x99), s(y82)) → EQ(x99, y82)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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:

EQ(s(x99), s(y82)) → EQ(x99, y82)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(77) Obligation:

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

EQ(s(x99), s(y82)) → EQ(x99, y82)

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:

• EQ(s(x99), s(y82)) → EQ(x99, y82)
  The graph contains the following edges 1 > 1, 2 > 2

(79) TRUE

(80) Obligation:

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

IF2(false, x125, y104, xs40) → DEL(x125, xs40)
DEL(x49, cons(y40, z7)) → IF2(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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:

IF2(false, x125, y104, xs40) → DEL(x125, xs40)
DEL(x49, cons(y40, z7)) → IF2(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(84) Obligation:

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

IF2(false, x125, y104, xs40) → DEL(x125, xs40)
DEL(x49, cons(y40, z7)) → IF2(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)

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(x49, cons(y40, z7)) → IF2(eq(x49, y40), x49, y40, z7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(false, x125, y104, xs40) → DEL(x125, xs40)
  The graph contains the following edges 2 >= 1, 4 >= 2

(86) TRUE

(87) Obligation:

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

MIN(x23, cons(y18, z2)) → IF1(le(x23, y18), x23, y18, z2)
IF1(true, x10, y7, xs1) → MIN(x10, xs1)
IF1(false, x36, y29, xs10) → MIN(y29, xs10)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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

(88) 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.

(89) Obligation:

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

MIN(x23, cons(y18, z2)) → IF1(le(x23, y18), x23, y18, z2)
IF1(true, x10, y7, xs1) → MIN(x10, xs1)
IF1(false, x36, y29, xs10) → MIN(y29, xs10)

The TRS R consists of the following rules:

le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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

(90) 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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, 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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(91) Obligation:

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

MIN(x23, cons(y18, z2)) → IF1(le(x23, y18), x23, y18, z2)
IF1(true, x10, y7, xs1) → MIN(x10, xs1)
IF1(false, x36, y29, xs10) → MIN(y29, xs10)

The TRS R consists of the following rules:

le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(92) 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:

• MIN(x23, cons(y18, z2)) → IF1(le(x23, y18), x23, y18, z2)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF1(true, x10, y7, xs1) → MIN(x10, xs1)
  The graph contains the following edges 2 >= 1, 4 >= 2

• IF1(false, x36, y29, xs10) → MIN(y29, xs10)
  The graph contains the following edges 3 >= 1, 4 >= 2

(93) TRUE

(94) Obligation:

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

IF2'(false, x125, y104, xs40) → DEL'(x125, xs40)
DEL'(x49, cons(y40, z7)) → IF2'(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

del'(x49, cons(y40, z7)) → if2'(eq(x49, y40), x49, y40, z7)
if2'(true, x112, y93, xs35) → true
if2'(false, x125, y104, xs40) → del'(x125, xs40)
del'(x138, nil) → false
min(x, nil) → x
if1(true, x10, y7, xs1) → min(x10, xs1)
min(x23, cons(y18, z2)) → if1(le(x23, y18), x23, y18, z2)
if1(false, x36, y29, xs10) → min(y29, xs10)
del(x49, cons(y40, z7)) → if2(eq(x49, y40), x49, y40, z7)
eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)
if2(true, x112, y93, xs35) → xs35
if2(false, x125, y104, xs40) → cons(y104, del(x125, xs40))
del(x138, nil) → nil
le(0, y125) → true
le(s(x163), 0) → false
le(s(x176), s(y146)) → le(x176, y146)
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[a32](nil, nil) → true
equal_sort[a32](nil, cons(x0, x1)) → false
equal_sort[a32](cons(x0, x1), nil) → false
equal_sort[a32](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a32](x0, x2), equal_sort[a32](x1, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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

(95) 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.

(96) Obligation:

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

IF2'(false, x125, y104, xs40) → DEL'(x125, xs40)
DEL'(x49, cons(y40, z7)) → IF2'(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

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

(97) 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))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
if1(true, x0, x1, x2)
min(x0, cons(x1, x2))
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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[a32](nil, nil)
equal_sort[a32](nil, cons(x0, x1))
equal_sort[a32](cons(x0, x1), nil)
equal_sort[a32](cons(x0, x1), cons(x2, x3))
equal_sort[a61](witness_sort[a61], witness_sort[a61])

(98) Obligation:

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

IF2'(false, x125, y104, xs40) → DEL'(x125, xs40)
DEL'(x49, cons(y40, z7)) → IF2'(eq(x49, y40), x49, y40, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y61)) → false
eq(s(x86), 0) → false
eq(s(x99), s(y82)) → eq(x99, y82)

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.

(99) 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'(x49, cons(y40, z7)) → IF2'(eq(x49, y40), x49, y40, z7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2'(false, x125, y104, xs40) → DEL'(x125, xs40)
  The graph contains the following edges 2 >= 1, 4 >= 2

(100) TRUE