(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) 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.
(9) 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.