(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The TRS R 2 is
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(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:
F(true, x) → F(eq(0, minus(x, x)), plus1(x))
F(true, x) → EQ(0, minus(x, x))
F(true, x) → MINUS(x, x)
F(true, x) → PLUS1(x)
PLUS1(x) → PLUS(s(0), x)
PLUS(s(x), y) → PLUS(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
EQ(s(x), s(y)) → EQ(x, y)
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(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 4 SCCs with 4 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:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(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:
EQ(s(x), s(y)) → EQ(x, y)
R is empty.
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(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)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(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) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(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:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(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)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(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:
- MINUS(s(x), s(y)) → MINUS(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:
PLUS(s(x), y) → PLUS(x, y)
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
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:
PLUS(s(x), y) → PLUS(x, y)
R is empty.
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
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].
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(25) 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.
(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:
- PLUS(s(x), y) → PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(27) YES
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x) → F(eq(0, minus(x, x)), plus1(x))
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
plus1(x0)
plus(0, x0)
plus(s(x0), x1)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(29) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x) → F(eq(0, minus(x, x)), plus1(x))
The TRS R consists of the following rules:
f(true, x) → f(eq(0, minus(x, x)), plus1(x))
plus1(x) → plus(s(0), x)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
Q is empty.
We have to consider all (P,Q,R)-chains.
(31) 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' = [ ] on the rule
F(
true,
s(
zr0))[
zr0 /
s(
zr0)]
n[
zr0 /
0] →
F(
true,
s(
s(
zr0)))[
zr0 /
s(
zr0)]
n[
zr0 /
0]
This rule is correct for the QDP as the following derivation shows:
F(
true,
s(
zr0))[
zr0 /
s(
zr0)]
n[
zr0 /
0] →
F(
true,
s(
s(
zr0)))[
zr0 /
s(
zr0)]
n[
zr0 /
0]
by Equivalence by Domain Renaming of the lhs with [
zl0 /
zr0]
intermediate steps: Equiv DR (rhs) - Equiv DR (lhs)
F(
true,
s(
zl1))[
zl1 /
s(
zl1)]
n[
zl1 /
0] →
F(
true,
s(
s(
zr1)))[
zr1 /
s(
zr1)]
n[
zr1 /
0]
by Rewrite t with the rewrite sequence
: [([0],eq(0, 0) -> true), ([1],plus1(x) -> plus(s(0), x)), ([1],plus(s(x), y) -> s(plus(x, y))), ([1,0],plus(0, y) -> y)]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, s(zl1))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0] → F(eq(0, 0), plus1(s(zr1)))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0]
by Narrowing at position: [0,1]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
F(true, s(zs2))[zs2 / s(zs2)]n[zs2 / y1] → F(eq(0, minus(y1, y1)), plus1(s(zs2)))[zs2 / s(zs2)]n[zs2 / y1]
by Narrowing at position: [0,1]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
F(true, x)[ ]n[ ] → F(eq(0, minus(x, x)), plus1(x))[ ]n[ ]
by Rule from TRS P
intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Equiv DR (lhs) - Equiv Sµ (rhs) - Equiv IPS (lhs) - Equiv Sµ (lhs) - Instantiate mu - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
minus(s(x), s(y))[x / s(x), y / s(y)]n[ ] → minus(x, y)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
minus(s(x), s(y))[ ]n[ ] → minus(x, y)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Instantiation
minus(x, 0)[ ]n[ ] → x[ ]n[ ]
by Rule from TRS R
(32) NO