(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) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.
(4) Complex Obligation (AND)
(5) 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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
(7) YES
(8) 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) 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
(10) YES
(11) 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)
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:
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 minimal (P,Q,R)-chains.
(15) NonLoopProof (EQUIVALENT 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:
intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming)
F(true, s(zl1))[zl1 / s(zl1)]n[zl1 / 0] → F(true, s(s(zr1)))[zr1 / s(zr1)]n[zr1 / 0]
by Rewrite t
intermediate steps: Equivalent (Remove Unused)
F(true, s(zl1))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / 0, zr1 / 0] → F(eq(0, 0), plus1(s(zr1)))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / 0, zr1 / 0]
by Narrowing at position: [0,1]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
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 OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Simplify mu) - Equivalent (Remove Unused) - Equivalent (Simplify mu) - Instantiate mu - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
minus(s(x), s(y))[x / s(x), y / s(y)]n[ ] → minus(x, y)[ ]n[ ]
by PatternCreation I
minus(s(x), s(y))[ ]n[ ] → minus(x, y)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Instantiation
minus(x, 0)[ ]n[ ] → x[ ]n[ ]
by OriginalRule from TRS R
(16) NO