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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The TRS R 2 is

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

(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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(true, x, y) → GR(y, 0)
COND2(true, x, y) → P(x)
COND2(true, x, y) → P(y)
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(false, x, y) → AND(eq(x, y), gr(x, 0))
COND2(false, x, y) → EQ(x, y)
COND2(false, x, y) → GR(x, 0)
GR(s(x), s(y)) → GR(x, y)
EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 7 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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

(11) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(13) TRUE

(14) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

(18) Obligation:

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

GR(s(x), s(y)) → GR(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:

• GR(s(x), s(y)) → GR(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:

COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)

(25) Obligation:

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

COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(26) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y)) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, y0, 0) → COND2(false, p(y0), p(0))
COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0)))

(27) Obligation:

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

COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, y0, 0) → COND2(false, p(y0), p(0))
COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0)))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(28) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, y0, 0) → COND2(false, p(y0), p(0)) at position [2] we obtained the following new rules [LPAR04]:

COND2(true, y0, 0) → COND2(false, p(y0), 0)

(29) Obligation:

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

COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0)))
COND2(true, y0, 0) → COND2(false, p(y0), 0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(30) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

(31) Obligation:

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

COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, y0, 0) → COND2(false, p(y0), 0)
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(32) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND1(true, x, y) → COND2(gr(y, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND1(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))

(33) Obligation:

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

COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(true, y0, 0) → COND2(false, p(y0), 0)
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(34) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y) we obtained the following new rules [LPAR04]:

COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0)

(35) Obligation:

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

COND2(true, y0, 0) → COND2(false, p(y0), 0)
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(36) DependencyGraphProof (EQUIVALENT transformation)

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

(37) Complex Obligation (AND)

(38) Obligation:

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

COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), 0) → false
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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].

p(0)
p(s(x0))

(42) Obligation:

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

COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), 0) → false
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(43) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, y_0, 0) → COND1(and(eq(y_0, 0), gr(y_0, 0)), y_0, 0) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

(44) Obligation:

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), 0) → false
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(45) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0) at position [0,1] we obtained the following new rules [LPAR04]:

COND2(false, 0, 0) → COND1(and(true, false), 0, 0)

(46) Obligation:

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND2(false, 0, 0) → COND1(and(true, false), 0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), 0) → false
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) Obligation:

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

COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), 0) → false
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(49) 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.

(50) Obligation:

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

COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

eq(s(x), 0) → false
and(true, true) → true
and(false, x) → false
eq(0, 0) → true
and(x, false) → false
gr(s(x), 0) → true

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(51) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, s(x0), 0) → COND1(false, s(x0), 0)

(52) Obligation:

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND2(false, s(x0), 0) → COND1(false, s(x0), 0)

The TRS R consists of the following rules:

eq(s(x), 0) → false
and(true, true) → true
and(false, x) → false
eq(0, 0) → true
and(x, false) → false
gr(s(x), 0) → true

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(53) DependencyGraphProof (EQUIVALENT transformation)

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

(54) Obligation:

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

COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

eq(s(x), 0) → false
and(true, true) → true
and(false, x) → false
eq(0, 0) → true
and(x, false) → false
gr(s(x), 0) → true

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(55) 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.

(56) Obligation:

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

COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(57) 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].

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

(58) Obligation:

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

COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(59) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, 0, 0) → COND1(false, 0, 0)

(60) Obligation:

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND2(false, 0, 0) → COND1(false, 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(61) DependencyGraphProof (EQUIVALENT transformation)

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

(62) Obligation:

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

COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(63) 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.

(64) Obligation:

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

COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(65) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0) at position [0,0] we obtained the following new rules [LPAR04]:

COND2(false, s(x0), 0) → COND1(and(false, true), s(x0), 0)

(66) Obligation:

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(and(false, true), s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(67) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(68) TRUE

(69) Obligation:

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(70) 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.

(71) Obligation:

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(72) 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].

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

(73) Obligation:

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(74) 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:

• COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
  The graph contains the following edges 1 >= 1, 3 > 3

(75) TRUE