(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The TRS R 2 is
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), 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:
F(true, x, y) → F(gt(x, y), plus2(x), plus1(y))
F(true, x, y) → GT(x, y)
F(true, x, y) → PLUS2(x)
F(true, x, y) → PLUS1(y)
GT(s(x), s(y)) → GT(x, y)
PLUS1(x) → PLUS(s(0), x)
PLUS2(x) → PLUS(s(s(0)), x)
PLUS(s(x), y) → PLUS(x, y)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), 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 5 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:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
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:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
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].
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
(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) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(x), s(y)) → GT(x, y)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
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:
GT(s(x), s(y)) → GT(x, y)
R is empty.
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
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].
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(x), s(y)) → GT(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:
- GT(s(x), s(y)) → GT(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(20) YES
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), plus2(x), plus1(y))
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
f(true, x0, x1)
gt(s(x0), 0)
gt(0, x0)
gt(s(x0), s(x1))
plus1(x0)
plus2(x0)
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(22) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), plus2(x), plus1(y))
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), plus2(x), plus1(y))
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
plus1(x) → plus(s(0), x)
plus2(x) → plus(s(s(0)), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(24) NonLoopProof (COMPLETE transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1,
σ' = [ ], and μ' = [
zr0 /
s(
zr0)] on the rule
F(
true,
s(
s(
s(
zr0))),
s(
zr1))[
zr0 /
s(
zr0),
zr1 /
s(
zr1)]
n[
zr1 /
0] →
F(
true,
s(
s(
s(
s(
zr0)))),
s(
s(
zr1)))[
zr0 /
s(
zr0),
zr1 /
s(
zr1)]
n[
zr1 /
0,
zr0 /
s(
zr0)]
This rule is correct for the QDP as the following derivation shows:
F(
true,
s(
s(
s(
zr0))),
s(
zr1))[
zr0 /
s(
zr0),
zr1 /
s(
zr1)]
n[
zr1 /
0] →
F(
true,
s(
s(
s(
s(
zr0)))),
s(
s(
zr1)))[
zr0 /
s(
zr0),
zr1 /
s(
zr1)]
n[
zr1 /
0,
zr0 /
s(
zr0)]
by Equivalency by Simplifying Mu with µ1: [
zr0 /
s(
zr0)] µ2: [
zr1 /
0]
intermediate steps: Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
F(
true,
s(
s(
zl2)),
s(
zl3))[
zl2 /
s(
zl2),
zl3 /
s(
zl3)]
n[
zl2 /
y0,
zl3 /
0] →
F(
true,
s(
s(
s(
s(
zr2)))),
s(
s(
zr3)))[
zr2 /
s(
zr2),
zr3 /
s(
zr3)]
n[
zr2 /
y0,
zr3 /
0]
by Rewrite t with the rewrite sequence
: [([1],plus2(x) -> plus(s(s(0)), x)), ([1],plus(s(x), y) -> s(plus(x, y))), ([1,0],plus(s(x), y) -> s(plus(x, y))), ([1,0,0],plus(0, y) -> y), ([2],plus1(x) -> plus(s(0), x)), ([2],plus(s(x), y) -> s(plus(x, y))), ([2,0],plus(0, y) -> y)]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, s(s(zl2)), s(zl3))[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / y0, zr3 / 0, zl2 / y0, zl3 / 0] → F(true, plus2(s(s(zr2))), plus1(s(zr3)))[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / y0, zr3 / 0, zl2 / y0, zl3 / 0]
by Narrowing at position: [0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv Sµ (rhs) - Equiv Sµ (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
F(true, s(zs2), s(zs3))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0] → F(gt(y1, y0), plus2(s(zs2)), plus1(s(zs3)))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation - Instantiation
F(true, x, y)[ ]n[ ] → F(gt(x, y), plus2(x), plus1(y))[ ]n[ ]
by Rule from TRS P
intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
gt(s(x), s(y))[x / s(x), y / s(y)]n[ ] → gt(x, y)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
gt(s(x), s(y))[ ]n[ ] → gt(x, y)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation
gt(s(x), 0)[ ]n[ ] → true[ ]n[ ]
by Rule from TRS R
(25) NO