(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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, s(x)) → F(eq(0, minus(x, x)), double(s(x)))
F(true, s(x)) → EQ(0, minus(x, x))
F(true, s(x)) → MINUS(x, x)
F(true, s(x)) → DOUBLE(s(x))
DOUBLE(x) → DOUBLEAKK(x, 0)
DOUBLEAKK(s(x), y) → DOUBLEAKK(x, s(s(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, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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:
DOUBLEAKK(s(x), y) → DOUBLEAKK(x, s(s(y)))
The TRS R consists of the following rules:
f(true, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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:
- DOUBLEAKK(s(x), y) → DOUBLEAKK(x, s(s(y)))
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, s(x)) → F(eq(0, minus(x, x)), double(s(x)))
The TRS R consists of the following rules:
f(true, s(x)) → f(eq(0, minus(x, x)), double(s(x)))
double(x) → doubleAkk(x, 0)
doubleAkk(0, y) → y
doubleAkk(s(x), y) → doubleAkk(x, s(s(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 = 2, b = 2,
σ' = [ ], and μ' = [ ] on the rule
F(true, s(s(zr0)))[zr0 / s(zr0)]n[zr0 / 0] → F(true, s(s(s(s(zr0)))))[zr0 / s(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(s(zl1)))[zl1 / s(zl1)]n[zl1 / 0] → F(true, s(s(s(s(zr1)))))[zr1 / s(s(zr1))]n[zr1 / 0]
by Rewrite t
intermediate steps: Equivalent (Remove Unused)
F(true, s(s(zl1)))[zl1 / s(zl1), zr1 / s(s(zr1))]n[zl1 / 0, zr1 / 0] → F(eq(0, minus(0, 0)), s(s(s(s(zr1)))))[zl1 / s(zl1), zr1 / s(s(zr1))]n[zl1 / 0, zr1 / 0]
by Narrowing at position: [1]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Simplify mu) - Equivalent (Remove Unused)
F(true, s(s(zl1)))[zl1 / s(zl1), zr1 / s(zr1), zt1 / s(s(zt1))]n[zl1 / y1, x0 / y1, zr1 / y1, zt1 / 0] → F(eq(0, minus(x0, x0)), doubleAkk(y1, s(s(s(s(zt1))))))[zl1 / s(zl1), zr1 / s(zr1), zt1 / s(s(zt1))]n[zl1 / y1, x0 / y1, zr1 / y1, zt1 / 0]
by Narrowing at position: [1]
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(true, s(s(zl1)))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / x0, zr1 / x0] → F(eq(0, minus(x0, x0)), doubleAkk(s(s(zr1)), 0))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / x0, zr1 / x0]
by Narrowing at position: [1]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
F(true, s(s(zs2)))[zs2 / s(zs2)]n[zs2 / y1] → F(eq(0, minus(y1, y1)), double(s(s(zs2))))[zs2 / s(zs2)]n[zs2 / y1]
by Narrowing at position: [0,1]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
F(true, s(x))[ ]n[ ] → F(eq(0, minus(x, x)), double(s(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) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation - Instantiation
double(x)[ ]n[ ] → doubleAkk(x, 0)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Expand Sigma - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
doubleAkk(s(x), y)[x / s(x)]n[ ] → doubleAkk(x, s(s(y)))[y / s(s(y))]n[ ]
by PatternCreation I
doubleAkk(s(x), y)[ ]n[ ] → doubleAkk(x, s(s(y)))[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation - Instantiation
doubleAkk(0, y)[ ]n[ ] → y[ ]n[ ]
by OriginalRule from TRS R
(16) NO