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:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

(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:

TABLE → GEN(s(0))
GEN(x) → IF1(le(x, 10), x)
GEN(x) → LE(x, 10)
GEN(x) → 10¹
IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
IF2(x, y) → LE(y, 10)
IF2(x, y) → 10¹
IF3(true, x, y) → TIMES(x, y)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
LE(s(x), s(y)) → LE(x, y)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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:

PLUS(s(x), y) → PLUS(x, y)

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

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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].

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

(11) Obligation:

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

PLUS(s(x), y) → PLUS(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:

• PLUS(s(x), y) → PLUS(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:

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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:

TIMES(s(x), y) → TIMES(x, y)

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

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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].

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

(18) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

R is empty.
Q is empty.
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:

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

(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:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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].

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

(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:

IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, 10), x)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

The set Q consists of the following terms:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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:

IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, 10), x)

The TRS R consists of the following rules:

10 → s(s(s(s(s(s(s(s(s(s(0))))))))))
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:

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
10

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].

table
gen(x0)
if1(false, x0)
if1(true, x0)
if2(x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

(32) Obligation:

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

IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, 10), x)

The TRS R consists of the following rules:

10 → s(s(s(s(s(s(s(s(s(s(0))))))))))
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))
10

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(x, y) → IF3(le(y, 10), x, y) we obtained the following new rules [LPAR04]:

IF2(z0, z0) → IF3(le(z0, 10), z0, z0)
IF2(z0, s(z1)) → IF3(le(s(z1), 10), z0, s(z1))

(34) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, 10), x)
IF2(z0, z0) → IF3(le(z0, 10), z0, z0)
IF2(z0, s(z1)) → IF3(le(s(z1), 10), z0, s(z1))

The TRS R consists of the following rules:

10 → s(s(s(s(s(s(s(s(s(s(0))))))))))
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))
10

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(x, y) → IF3(le(y, 10), x, y) at position [0,1] we obtained the following new rules [LPAR04]:

IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)

(36) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, 10), x)
IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)

The TRS R consists of the following rules:

10 → s(s(s(s(s(s(s(s(s(s(0))))))))))
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))
10

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

(37) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule GEN(x) → IF1(le(x, 10), x) at position [0,1] we obtained the following new rules [LPAR04]:

GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

(38) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)
GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

10 → s(s(s(s(s(s(s(s(s(s(0))))))))))
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))
10

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

(39) 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.

(40) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)
GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
10

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

(41) 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].

10

(42) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)
GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

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

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.

(43) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y) we obtained the following new rules [LPAR04]:

IF2(z0, z0) → IF3(le(z0, s(s(s(s(s(s(s(s(s(s(0))))))))))), z0, z0)
IF2(z0, s(z1)) → IF3(le(s(z1), s(s(s(s(s(s(s(s(s(s(0))))))))))), z0, s(z1))

(44) Obligation:

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

IF1(true, x) → IF2(x, x)
IF3(true, x, y) → IF2(x, s(y))
IF3(false, x, y) → GEN(s(x))
GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)
IF2(z0, z0) → IF3(le(z0, s(s(s(s(s(s(s(s(s(s(0))))))))))), z0, z0)
IF2(z0, s(z1)) → IF3(le(s(z1), s(s(s(s(s(s(s(s(s(s(0))))))))))), z0, s(z1))

The TRS R consists of the following rules:

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

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.

(45) RemovalProof (SOUND transformation)

In the following pairs the term without variables s(s(s(s(s(s(s(s(s(s(0)))))))))) is replaced by the fresh variable x_removed.
Pair: IF2(x, y) → IF3(le(y, s(s(s(s(s(s(s(s(s(s(0))))))))))), x, y)
Positions in right side of the pair:

• [0,1]

Pair: GEN(x) → IF1(le(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)
Positions in right side of the pair:

• [0,1]

The new variable was added to all pairs as a new argument[CONREM].

(46) Obligation:

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

IF1(true, x, x_removed) → IF2(x, x, x_removed)
IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF3(false, x, y, x_removed) → GEN(s(x), x_removed)
GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

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.

(47) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF1(true, x, x_removed) → IF2(x, x, x_removed) the following chains were created:

• We consider the chain GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed), IF1(true, x, x_removed) → IF2(x, x, x_removed) which results in the following constraint:
  

  (1)    (IF1(le(x11, x12), x11, x12)=IF1(true, x13, x14) ⇒ IF1(true, x13, x14)≥IF2(x13, x13, x14))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (le(x11, x12)=true ⇒ IF1(true, x11, x12)≥IF2(x11, x11, x12))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x11, x12)=true which results in the following new constraints:
  

  (3)    (true=true ⇒ IF1(true, 0, x81)≥IF2(0, 0, x81))

  
  

  (4)    (le(x83, x82)=true∧(le(x83, x82)=true ⇒ IF1(true, x83, x82)≥IF2(x83, x83, x82)) ⇒ IF1(true, s(x83), s(x82))≥IF2(s(x83), s(x83), s(x82)))

  
  
  We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
  

  (5)    (IF1(true, 0, x81)≥IF2(0, 0, x81))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (le(x83, x82)=true ⇒ IF1(true, x83, x82)≥IF2(x83, x83, x82)) with σ = [ ] which results in the following new constraint:
  

  (6)    (IF1(true, x83, x82)≥IF2(x83, x83, x82) ⇒ IF1(true, s(x83), s(x82))≥IF2(s(x83), s(x83), s(x82)))

  
  
  

For Pair IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) the following chains were created:

• We consider the chain IF1(true, x, x_removed) → IF2(x, x, x_removed), IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) which results in the following constraint:
  

  (7)    (IF2(x15, x15, x16)=IF2(x17, x18, x19) ⇒ IF2(x17, x18, x19)≥IF3(le(x18, x19), x17, x18, x19))

  
  
  We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:
  

  (8)    (IF2(x15, x15, x16)≥IF3(le(x15, x16), x15, x15, x16))

  
  
  
• We consider the chain IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed), IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) which results in the following constraint:
  

  (9)    (IF2(x23, s(x24), x25)=IF2(x26, x27, x28) ⇒ IF2(x26, x27, x28)≥IF3(le(x27, x28), x26, x27, x28))

  
  
  We simplified constraint (9) using rules (I), (II), (III) which results in the following new constraint:
  

  (10)    (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))

  
  
  

For Pair IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) the following chains were created:

• We consider the chain IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed), IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) which results in the following constraint:
  

  (11)    (IF3(le(x37, x38), x36, x37, x38)=IF3(true, x39, x40, x41) ⇒ IF3(true, x39, x40, x41)≥IF2(x39, s(x40), x41))

  
  
  We simplified constraint (11) using rules (I), (II), (III) which results in the following new constraint:
  

  (12)    (le(x37, x38)=true ⇒ IF3(true, x36, x37, x38)≥IF2(x36, s(x37), x38))

  
  
  We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on le(x37, x38)=true which results in the following new constraints:
  

  (13)    (true=true ⇒ IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  

  (14)    (le(x87, x86)=true∧(∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  We simplified constraint (13) using rules (I), (II) which results in the following new constraint:
  

  (15)    (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  
  We simplified constraint (14) using rule (VI) where we applied the induction hypothesis (∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) with σ = [x88 / x36] which results in the following new constraint:
  

  (16)    (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  

For Pair IF3(false, x, y, x_removed) → GEN(s(x), x_removed) the following chains were created:

• We consider the chain IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed), IF3(false, x, y, x_removed) → GEN(s(x), x_removed) which results in the following constraint:
  

  (17)    (IF3(le(x53, x54), x52, x53, x54)=IF3(false, x55, x56, x57) ⇒ IF3(false, x55, x56, x57)≥GEN(s(x55), x57))

  
  
  We simplified constraint (17) using rules (I), (II), (III) which results in the following new constraint:
  

  (18)    (le(x53, x54)=false ⇒ IF3(false, x52, x53, x54)≥GEN(s(x52), x54))

  
  
  We simplified constraint (18) using rule (V) (with possible (I) afterwards) using induction on le(x53, x54)=false which results in the following new constraints:
  

  (19)    (le(x92, x91)=false∧(∀x93:le(x92, x91)=false ⇒ IF3(false, x93, x92, x91)≥GEN(s(x93), x91)) ⇒ IF3(false, x52, s(x92), s(x91))≥GEN(s(x52), s(x91)))

  
  

  (20)    (false=false ⇒ IF3(false, x52, s(x94), 0)≥GEN(s(x52), 0))

  
  
  We simplified constraint (19) using rule (VI) where we applied the induction hypothesis (∀x93:le(x92, x91)=false ⇒ IF3(false, x93, x92, x91)≥GEN(s(x93), x91)) with σ = [x93 / x52] which results in the following new constraint:
  

  (21)    (IF3(false, x52, x92, x91)≥GEN(s(x52), x91) ⇒ IF3(false, x52, s(x92), s(x91))≥GEN(s(x52), s(x91)))

  
  
  We simplified constraint (20) using rules (I), (II) which results in the following new constraint:
  

  (22)    (IF3(false, x52, s(x94), 0)≥GEN(s(x52), 0))

  
  
  

For Pair GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed) the following chains were created:

• We consider the chain IF3(false, x, y, x_removed) → GEN(s(x), x_removed), GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed) which results in the following constraint:
  

  (23)    (GEN(s(x74), x76)=GEN(x77, x78) ⇒ GEN(x77, x78)≥IF1(le(x77, x78), x77, x78))

  
  
  We simplified constraint (23) using rules (I), (II), (III) which results in the following new constraint:
  

  (24)    (GEN(s(x74), x76)≥IF1(le(s(x74), x76), s(x74), x76))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF1(true, x, x_removed) → IF2(x, x, x_removed)
  
  • (IF1(true, 0, x81)≥IF2(0, 0, x81))
    
  • (IF1(true, x83, x82)≥IF2(x83, x83, x82) ⇒ IF1(true, s(x83), s(x82))≥IF2(s(x83), s(x83), s(x82)))
    
  
  
• IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
  
  • (IF2(x15, x15, x16)≥IF3(le(x15, x16), x15, x15, x16))
    
  • (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))
    
  
  
• IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
  
  • (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))
    
  • (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))
    
  
  
• IF3(false, x, y, x_removed) → GEN(s(x), x_removed)
  
  • (IF3(false, x52, x92, x91)≥GEN(s(x52), x91) ⇒ IF3(false, x52, s(x92), s(x91))≥GEN(s(x52), s(x91)))
    
  • (IF3(false, x52, s(x94), 0)≥GEN(s(x52), 0))
    
  
  
• GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed)
  
  • (GEN(s(x74), x76)≥IF1(le(s(x74), x76), s(x74), x76))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(GEN(x₁, x₂)) = -x₁ + x₂   
POL(IF1(x₁, x₂, x₃)) = -x₁ - x₂ + x₃   
POL(IF2(x₁, x₂, x₃)) = -x₁ + x₃   
POL(IF3(x₁, x₂, x₃, x₄)) = -x₁ - x₂ + x₄   
POL(c) = -1   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF3(false, x, y, x_removed) → GEN(s(x), x_removed)

The following pairs are in P_bound:

IF1(true, x, x_removed) → IF2(x, x, x_removed)

The following rules are usable:

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

(48) Complex Obligation (AND)

(49) Obligation:

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

IF1(true, x, x_removed) → IF2(x, x, x_removed)
IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

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.

(50) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(51) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The TRS R consists of the following rules:

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

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.

(52) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) we obtained the following new rules [LPAR04]:

IF2(z0, s(z1), z2) → IF3(le(s(z1), z2), z0, s(z1), z2)

(53) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF2(z0, s(z1), z2) → IF3(le(s(z1), z2), z0, s(z1), z2)

The TRS R consists of the following rules:

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

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.

(54) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) the following chains were created:

• We consider the chain IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed), IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) which results in the following constraint:
  

  (1)    (IF2(x23, s(x24), x25)=IF2(x26, x27, x28) ⇒ IF2(x26, x27, x28)≥IF3(le(x27, x28), x26, x27, x28))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))

  
  
  

For Pair IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) the following chains were created:

• We consider the chain IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed), IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) which results in the following constraint:
  

  (3)    (IF3(le(x37, x38), x36, x37, x38)=IF3(true, x39, x40, x41) ⇒ IF3(true, x39, x40, x41)≥IF2(x39, s(x40), x41))

  
  
  We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
  

  (4)    (le(x37, x38)=true ⇒ IF3(true, x36, x37, x38)≥IF2(x36, s(x37), x38))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on le(x37, x38)=true which results in the following new constraints:
  

  (5)    (true=true ⇒ IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  

  (6)    (le(x87, x86)=true∧(∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
  

  (7)    (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  
  We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) with σ = [x88 / x36] which results in the following new constraint:
  

  (8)    (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
  
  • (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))
    
  
  
• IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
  
  • (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))
    
  • (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(IF2(x₁, x₂, x₃)) = 1 + x₁ - x₂ + x₃   
POL(IF3(x₁, x₂, x₃, x₄)) = -x₁ + x₂ - x₃ + x₄   
POL(c) = -1   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The following pairs are in P_bound:

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)

The following rules are usable:

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

(55) Complex Obligation (AND)

(56) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)

The TRS R consists of the following rules:

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

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.

(57) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(58) TRUE

(59) Obligation:

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

IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The TRS R consists of the following rules:

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

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.

(60) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(61) TRUE

(62) Obligation:

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

IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF3(false, x, y, x_removed) → GEN(s(x), x_removed)
GEN(x, x_removed) → IF1(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

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.

(63) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(64) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The TRS R consists of the following rules:

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

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.

(65) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) we obtained the following new rules [LPAR04]:

IF2(z0, s(z1), z2) → IF3(le(s(z1), z2), z0, s(z1), z2)

(66) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
IF2(z0, s(z1), z2) → IF3(le(s(z1), z2), z0, s(z1), z2)

The TRS R consists of the following rules:

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

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.

(67) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) the following chains were created:

• We consider the chain IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed), IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed) which results in the following constraint:
  

  (1)    (IF2(x23, s(x24), x25)=IF2(x26, x27, x28) ⇒ IF2(x26, x27, x28)≥IF3(le(x27, x28), x26, x27, x28))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))

  
  
  

For Pair IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) the following chains were created:

• We consider the chain IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed), IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed) which results in the following constraint:
  

  (3)    (IF3(le(x37, x38), x36, x37, x38)=IF3(true, x39, x40, x41) ⇒ IF3(true, x39, x40, x41)≥IF2(x39, s(x40), x41))

  
  
  We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
  

  (4)    (le(x37, x38)=true ⇒ IF3(true, x36, x37, x38)≥IF2(x36, s(x37), x38))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on le(x37, x38)=true which results in the following new constraints:
  

  (5)    (true=true ⇒ IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  

  (6)    (le(x87, x86)=true∧(∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
  

  (7)    (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))

  
  
  We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x88:le(x87, x86)=true ⇒ IF3(true, x88, x87, x86)≥IF2(x88, s(x87), x86)) with σ = [x88 / x36] which results in the following new constraint:
  

  (8)    (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)
  
  • (IF2(x23, s(x24), x25)≥IF3(le(s(x24), x25), x23, s(x24), x25))
    
  
  
• IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)
  
  • (IF3(true, x36, 0, x85)≥IF2(x36, s(0), x85))
    
  • (IF3(true, x36, x87, x86)≥IF2(x36, s(x87), x86) ⇒ IF3(true, x36, s(x87), s(x86))≥IF2(x36, s(s(x87)), s(x86)))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(IF2(x₁, x₂, x₃)) = 1 + x₁ - x₂ + x₃   
POL(IF3(x₁, x₂, x₃, x₄)) = -x₁ + x₂ - x₃ + x₄   
POL(c) = -1   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The following pairs are in P_bound:

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)

The following rules are usable:

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

(68) Complex Obligation (AND)

(69) Obligation:

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

IF3(true, x, y, x_removed) → IF2(x, s(y), x_removed)

The TRS R consists of the following rules:

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

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.

(70) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(71) TRUE

(72) Obligation:

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

IF2(x, y, x_removed) → IF3(le(y, x_removed), x, y, x_removed)

The TRS R consists of the following rules:

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

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.

(73) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(74) TRUE