(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