(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
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:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
The set Q consists of the following terms:
diff(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
(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:
DIFF(x, y) → COND1(equal(x, y), x, y)
DIFF(x, y) → EQUAL(x, y)
COND1(false, x, y) → COND2(gt(x, y), x, y)
COND1(false, x, y) → GT(x, y)
COND2(true, x, y) → DIFF(x, s(y))
COND2(false, x, y) → DIFF(s(x), y)
GT(s(u), s(v)) → GT(u, v)
EQUAL(s(x), s(y)) → EQUAL(x, y)
The TRS R consists of the following rules:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
The set Q consists of the following terms:
diff(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
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 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQUAL(s(x), s(y)) → EQUAL(x, y)
The TRS R consists of the following rules:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
The set Q consists of the following terms:
diff(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
The set Q consists of the following terms:
diff(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(false, x, y) → COND2(gt(x, y), x, y)
COND2(true, x, y) → DIFF(x, s(y))
DIFF(x, y) → COND1(equal(x, y), x, y)
COND2(false, x, y) → DIFF(s(x), y)
The TRS R consists of the following rules:
diff(x, y) → cond1(equal(x, y), x, y)
cond1(true, x, y) → 0
cond1(false, x, y) → cond2(gt(x, y), x, y)
cond2(true, x, y) → s(diff(x, s(y)))
cond2(false, x, y) → s(diff(s(x), y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
The set Q consists of the following terms:
diff(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.