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) YES

(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) YES

(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) YES

(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) YES

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

At least one of these decreasing rules is always used after the deleted DP:
if2(true, x1137, y947, xs367) → xs367


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


The transformed set:
del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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 proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, x113, y94, xs36, x126, y105, y41, z7, x139, x50, x11, y8, z'', y62, x87, x100, y83, y126, x164, x177, y147, x24, y19, x37, y30]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or 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)) -> equal_sort[a32](x0, x2) and equal_sort[a32](x1, x3)
   equal_sort[a61](witness_sort[a61], witness_sort[a61]) -> true
   if2'(true, x113, y94, xs36) -> true
   if2'(false, x126, y105, cons(y41, z7)) -> if2'(eq(x126, y41), x126, y41, z7)
   if2'(false, x126, y105, nil) -> false
   del'(x139, nil) -> false
   equal_bool(eq(x50, y41), true) -> true | del'(x50, cons(y41, z7)) -> true
   equal_bool(eq(x50, y41), true) -> false | del'(x50, cons(y41, z7)) -> del'(x50, z7)
   min(x, nil) -> x
   equal_bool(le(x11, y8), true) -> true | min(x11, cons(y8, z'')) -> min(x11, z'')
   equal_bool(le(x11, y8), true) -> false | min(x11, cons(y8, z'')) -> min(y8, z'')
   eq(0, 0) -> true
   eq(0, s(y62)) -> false
   eq(s(x87), 0) -> false
   eq(s(x100), s(y83)) -> eq(x100, y83)
   if2(true, x113, y94, xs36) -> xs36
   if2(false, x126, y105, cons(y41, z7)) -> cons(y105, if2(eq(x126, y41), x126, y41, z7))
   if2(false, x126, y105, nil) -> cons(y105, nil)
   del(x139, nil) -> nil
   equal_bool(eq(x50, y41), true) -> true | del(x50, cons(y41, z7)) -> z7
   equal_bool(eq(x50, y41), true) -> false | del(x50, cons(y41, z7)) -> cons(y41, del(x50, z7))
   le(0, y126) -> true
   le(s(x164), 0) -> false
   le(s(x177), s(y147)) -> le(x177, y147)
   if1(true, x24, y19, nil) -> x24
   if1(true, x24, y19, cons(y8, z'')) -> if1(le(x24, y8), x24, y8, z'')
   if1(false, x37, y30, nil) -> y30
   if1(false, x37, y30, cons(y8, z'')) -> if1(le(y30, y8), y30, y8, z'')

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, x113, y94, xs36, x126, y105, y41, z7, x139, x50, x11, y8, z'', y62, x87, x100, y83, y126, x164, x177, y147, x24, y19, x37, y30]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or 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)) -> equal_sort[a32](x0, x2) and equal_sort[a32](x1, x3)
   equal_sort[a61](witness_sort[a61], witness_sort[a61]) -> true
   if2'(true, x113, y94, xs36) -> true
   if2'(false, x126, y105, cons(y41, z7)) -> if2'(eq(x126, y41), x126, y41, z7)
   if2'(false, x126, y105, nil) -> false
   del'(x139, nil) -> false
   equal_bool(eq(x50, y41), true) -> true | del'(x50, cons(y41, z7)) -> true
   equal_bool(eq(x50, y41), true) -> false | del'(x50, cons(y41, z7)) -> del'(x50, z7)
   min(x, nil) -> x
   equal_bool(le(x11, y8), true) -> true | min(x11, cons(y8, z'')) -> min(x11, z'')
   equal_bool(le(x11, y8), true) -> false | min(x11, cons(y8, z'')) -> min(y8, z'')
   eq(0, 0) -> true
   eq(0, s(y62)) -> false
   eq(s(x87), 0) -> false
   eq(s(x100), s(y83)) -> eq(x100, y83)
   if2(true, x113, y94, xs36) -> xs36
   if2(false, x126, y105, cons(y41, z7)) -> cons(y105, if2(eq(x126, y41), x126, y41, z7))
   if2(false, x126, y105, nil) -> cons(y105, nil)
   del(x139, nil) -> nil
   equal_bool(eq(x50, y41), true) -> true | del(x50, cons(y41, z7)) -> z7
   equal_bool(eq(x50, y41), true) -> false | del(x50, cons(y41, z7)) -> cons(y41, del(x50, z7))
   le(0, y126) -> true
   le(s(x164), 0) -> false
   le(s(x177), s(y147)) -> le(x177, y147)
   if1(true, x24, y19, nil) -> x24
   if1(true, x24, y19, cons(y8, z'')) -> if1(le(x24, y8), x24, y8, z'')
   if1(false, x37, y30, nil) -> y30
   if1(false, x37, y30, cons(y8, z'')) -> if1(le(y30, y8), y30, y8, z'')

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

could be successfully shown:
(0) Formula
(1) Conditional Evaluation [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) YES
    (6) Formula
        (7) Hypothesis Lifting [EQUIVALENT]
        (8) Formula
        (9) Induction by algorithm [EQUIVALENT]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT]
                (13) Formula
                (14) Induction by data structure [EQUIVALENT]
                (15) AND
                    (16) Formula
                        (17) Symbolic evaluation [EQUIVALENT]
                        (18) YES
                    (19) Formula
                        (20) Symbolic evaluation under hypothesis [EQUIVALENT]
                        (21) YES
            (22) Formula
                (23) Conditional Evaluation [EQUIVALENT]
                (24) Formula
                (25) Conditional Evaluation [EQUIVALENT]
                (26) Formula
                (27) Conditional Evaluation [EQUIVALENT]
                (28) AND
                    (29) Formula
                        (30) Symbolic evaluation [EQUIVALENT]
                        (31) YES
                    (32) Formula
                        (33) Symbolic evaluation under hypothesis [EQUIVALENT]
                        (34) YES
            (35) Formula
                (36) Conditional Evaluation [EQUIVALENT]
                (37) Formula
                (38) Conditional Evaluation [EQUIVALENT]
                (39) Formula
                (40) Conditional Evaluation [EQUIVALENT]
                (41) AND
                    (42) Formula
                        (43) Symbolic evaluation [EQUIVALENT]
                        (44) YES
                    (45) Formula
                        (46) Symbolic evaluation under hypothesis [EQUIVALENT]
                        (47) YES


----------------------------------------

(0)
Obligation:
Formula:
x:sort[a0],y:sort[a32].del'(min(x, y), cons(x, y))=true

There are no hypotheses.




----------------------------------------

(1) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
x:sort[a0],y:sort[a32].equal_bool(eq(min(x, y), x), true)=true





Formula:
x:sort[a0],y:sort[a32].del'(min(x, y), y)=true

Hypotheses:
x:sort[a0],y:sort[a32].equal_bool(eq(min(x, y), x), true)=false






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
true=true

Hypotheses:
x:sort[a0],y:sort[a32].equal_bool(eq(min(x, y), x), true)=true




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

----------------------------------------

(6)
Obligation:
Formula:
x:sort[a0],y:sort[a32].del'(min(x, y), y)=true

Hypotheses:
x:sort[a0],y:sort[a32].equal_bool(eq(min(x, y), x), true)=false




----------------------------------------

(7) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x:sort[a0],y:sort[a32].(equal_bool(eq(min(x, y), x), true)=false->del'(min(x, y), y)=true)

There are no hypotheses.




----------------------------------------

(8)
Obligation:
Formula:
x:sort[a0],y:sort[a32].(equal_bool(eq(min(x, y), x), true)=false->del'(min(x, y), y)=true)

There are no hypotheses.




----------------------------------------

(9) Induction by algorithm (EQUIVALENT)
Induction by algorithm min(x, y) generates the following cases:

1. Base Case:
Formula:
x':sort[a0].(equal_bool(eq(min(x', nil), x'), true)=false->del'(min(x', nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x11:sort[a0],y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, cons(y8, z'')), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true





2. Step Case:
Formula:
x11:sort[a0],y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, cons(y8, z'')), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false






----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Formula:
x':sort[a0].(equal_bool(eq(min(x', nil), x'), true)=false->del'(min(x', nil), nil)=true)

There are no hypotheses.




----------------------------------------

(12) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x':sort[a0].~(equal_bool(eq(x', x'), true)=false)
----------------------------------------

(13)
Obligation:
Formula:
x':sort[a0].~(equal_bool(eq(x', x'), true)=false)

There are no hypotheses.




----------------------------------------

(14) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
~(equal_bool(eq(0, 0), true)=false)

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].~(equal_bool(eq(s(n), s(n)), true)=false)

Hypotheses:
n:sort[a0].~(equal_bool(eq(n, n), true)=false)






----------------------------------------

(15)
Complex Obligation (AND)

----------------------------------------

(16)
Obligation:
Formula:
~(equal_bool(eq(0, 0), true)=false)

There are no hypotheses.




----------------------------------------

(17) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(18)
YES

----------------------------------------

(19)
Obligation:
Formula:
n:sort[a0].~(equal_bool(eq(s(n), s(n)), true)=false)

Hypotheses:
n:sort[a0].~(equal_bool(eq(n, n), true)=false)




----------------------------------------

(20) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a0].~(equal_bool(eq(n, n), true)=false)

----------------------------------------

(21)
YES

----------------------------------------

(22)
Obligation:
Formula:
x11:sort[a0],y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, cons(y8, z'')), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true




----------------------------------------

(23) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x11:sort[a0],z'':sort[a32],y8:sort[a0].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true






----------------------------------------

(24)
Obligation:
Formula:
x11:sort[a0],z'':sort[a32],y8:sort[a0].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true




----------------------------------------

(25) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x11:sort[a0],z'':sort[a32],y8:sort[a0].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true






----------------------------------------

(26)
Obligation:
Formula:
x11:sort[a0],z'':sort[a32],y8:sort[a0].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), cons(y8, z''))=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true




----------------------------------------

(27) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->true=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true
x11:sort[a0],z'':sort[a32],y8:sort[a0].equal_bool(eq(min(x11, z''), y8), true)=true





Formula:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true
x11:sort[a0],z'':sort[a32],y8:sort[a0].equal_bool(eq(min(x11, z''), y8), true)=false






----------------------------------------

(28)
Complex Obligation (AND)

----------------------------------------

(29)
Obligation:
Formula:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->true=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true
x11:sort[a0],z'':sort[a32],y8:sort[a0].equal_bool(eq(min(x11, z''), y8), true)=true




----------------------------------------

(30) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(31)
YES

----------------------------------------

(32)
Obligation:
Formula:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)

Hypotheses:
x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=true
x11:sort[a0],z'':sort[a32],y8:sort[a0].equal_bool(eq(min(x11, z''), y8), true)=false




----------------------------------------

(33) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

x11:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, z''), x11), true)=false->del'(min(x11, z''), z'')=true)

----------------------------------------

(34)
YES

----------------------------------------

(35)
Obligation:
Formula:
x11:sort[a0],y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(x11, cons(y8, z'')), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false




----------------------------------------

(36) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false






----------------------------------------

(37)
Obligation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(x11, cons(y8, z'')), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false




----------------------------------------

(38) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(y8, z''), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false






----------------------------------------

(39)
Obligation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(y8, z''), cons(y8, z''))=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false




----------------------------------------

(40) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->true=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false
y8:sort[a0],z'':sort[a32].equal_bool(eq(min(y8, z''), y8), true)=true





Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(y8, z''), z'')=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false
y8:sort[a0],z'':sort[a32].equal_bool(eq(min(y8, z''), y8), true)=false






----------------------------------------

(41)
Complex Obligation (AND)

----------------------------------------

(42)
Obligation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->true=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false
y8:sort[a0],z'':sort[a32].equal_bool(eq(min(y8, z''), y8), true)=true




----------------------------------------

(43) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(44)
YES

----------------------------------------

(45)
Obligation:
Formula:
y8:sort[a0],z'':sort[a32],x11:sort[a0].(equal_bool(eq(min(y8, z''), x11), true)=false->del'(min(y8, z''), z'')=true)

Hypotheses:
y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
x11:sort[a0],y8:sort[a0].equal_bool(le(x11, y8), true)=false
y8:sort[a0],z'':sort[a32].equal_bool(eq(min(y8, z''), y8), true)=false




----------------------------------------

(46) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

y8:sort[a0],z'':sort[a32].(equal_bool(eq(min(y8, z''), y8), true)=false->del'(min(y8, z''), z'')=true)
y8:sort[a0],z'':sort[a32].equal_bool(eq(min(y8, z''), y8), true)=false

----------------------------------------

(47)
YES

(41) Complex Obligation (AND)

(42) Obligation:

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

min(x, nil) → x
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)
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) YES

(45) Obligation:

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

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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'(x50, cons(y41, z7)) → IF2'(eq(x50, y41), x50, y41, z7)
DEL'(x50, cons(y41, z7)) → EQ(x50, y41)
IF2'(false, x126, y105, xs41) → DEL'(x126, xs41)
MIN(x11, cons(y8, z'')) → IF1(le(x11, y8), x11, y8, z'')
MIN(x11, cons(y8, z'')) → LE(x11, y8)
IF1(true, x24, y19, xs6) → MIN(x24, xs6)
IF1(false, x37, y30, xs11) → MIN(y30, xs11)
DEL(x50, cons(y41, z7)) → IF2(eq(x50, y41), x50, y41, z7)
DEL(x50, cons(y41, z7)) → EQ(x50, y41)
EQ(s(x100), s(y83)) → EQ(x100, y83)
IF2(false, x126, y105, xs41) → DEL(x126, xs41)
LE(s(x177), s(y147)) → LE(x177, y147)
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'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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) YES

(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'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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) YES

(66) Obligation:

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

LE(s(x177), s(y147)) → LE(x177, y147)

The TRS R consists of the following rules:

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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(x177), s(y147)) → LE(x177, y147)

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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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(x177), s(y147)) → LE(x177, y147)

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(x177), s(y147)) → LE(x177, y147)
  The graph contains the following edges 1 > 1, 2 > 2

(72) YES

(73) Obligation:

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

EQ(s(x100), s(y83)) → EQ(x100, y83)

The TRS R consists of the following rules:

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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(x100), s(y83)) → EQ(x100, y83)

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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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(x100), s(y83)) → EQ(x100, y83)

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(x100), s(y83)) → EQ(x100, y83)
  The graph contains the following edges 1 > 1, 2 > 2

(79) YES

(80) Obligation:

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

IF2(false, x126, y105, xs41) → DEL(x126, xs41)
DEL(x50, cons(y41, z7)) → IF2(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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, x126, y105, xs41) → DEL(x126, xs41)
DEL(x50, cons(y41, z7)) → IF2(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)

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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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, x126, y105, xs41) → DEL(x126, xs41)
DEL(x50, cons(y41, z7)) → IF2(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)

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(x50, cons(y41, z7)) → IF2(eq(x50, y41), x50, y41, z7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(false, x126, y105, xs41) → DEL(x126, xs41)
  The graph contains the following edges 2 >= 1, 4 >= 2

(86) YES

(87) Obligation:

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

IF1(true, x24, y19, xs6) → MIN(x24, xs6)
MIN(x11, cons(y8, z'')) → IF1(le(x11, y8), x11, y8, z'')
IF1(false, x37, y30, xs11) → MIN(y30, xs11)

The TRS R consists of the following rules:

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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:

IF1(true, x24, y19, xs6) → MIN(x24, xs6)
MIN(x11, cons(y8, z'')) → IF1(le(x11, y8), x11, y8, z'')
IF1(false, x37, y30, xs11) → MIN(y30, xs11)

The TRS R consists of the following rules:

le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)

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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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:

IF1(true, x24, y19, xs6) → MIN(x24, xs6)
MIN(x11, cons(y8, z'')) → IF1(le(x11, y8), x11, y8, z'')
IF1(false, x37, y30, xs11) → MIN(y30, xs11)

The TRS R consists of the following rules:

le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)

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(x11, cons(y8, z'')) → IF1(le(x11, y8), x11, y8, z'')
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF1(true, x24, y19, xs6) → MIN(x24, xs6)
  The graph contains the following edges 2 >= 1, 4 >= 2

• IF1(false, x37, y30, xs11) → MIN(y30, xs11)
  The graph contains the following edges 3 >= 1, 4 >= 2

(93) YES

(94) Obligation:

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

IF2'(false, x126, y105, xs41) → DEL'(x126, xs41)
DEL'(x50, cons(y41, z7)) → IF2'(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

del'(x50, cons(y41, z7)) → if2'(eq(x50, y41), x50, y41, z7)
if2'(true, x113, y94, xs36) → true
if2'(false, x126, y105, xs41) → del'(x126, xs41)
del'(x139, nil) → false
min(x, nil) → x
min(x11, cons(y8, z'')) → if1(le(x11, y8), x11, y8, z'')
if1(true, x24, y19, xs6) → min(x24, xs6)
if1(false, x37, y30, xs11) → min(y30, xs11)
del(x50, cons(y41, z7)) → if2(eq(x50, y41), x50, y41, z7)
eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)
if2(true, x113, y94, xs36) → xs36
if2(false, x126, y105, xs41) → cons(y105, del(x126, xs41))
del(x139, nil) → nil
le(0, y126) → true
le(s(x164), 0) → false
le(s(x177), s(y147)) → le(x177, y147)
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)
min(x0, cons(x1, x2))
if1(true, x0, 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, x126, y105, xs41) → DEL'(x126, xs41)
DEL'(x50, cons(y41, z7)) → IF2'(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)

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)
min(x0, cons(x1, x2))
if1(true, x0, 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)
min(x0, cons(x1, x2))
if1(true, x0, 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, x126, y105, xs41) → DEL'(x126, xs41)
DEL'(x50, cons(y41, z7)) → IF2'(eq(x50, y41), x50, y41, z7)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y62)) → false
eq(s(x87), 0) → false
eq(s(x100), s(y83)) → eq(x100, y83)

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'(x50, cons(y41, z7)) → IF2'(eq(x50, y41), x50, y41, z7)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2'(false, x126, y105, xs41) → DEL'(x126, xs41)
  The graph contains the following edges 2 >= 1, 4 >= 2

(100) YES