NO Termination w.r.t. Q proof of AProVE_10_andIsNat.ari

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

The TRS R 2 is

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))

The signature Sigma is {f, cond}

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

The set Q consists of the following terms:

f(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

(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(x, y) → COND(and(isNat(x), isNat(y)), x, y)
F(x, y) → AND(isNat(x), isNat(y))
F(x, y) → ISNAT(x)
F(x, y) → ISNAT(y)
COND(tt, x, y) → F(s(x), s(y))
ISNAT(s(x)) → ISNAT(x)

The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

The set Q consists of the following terms:

f(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ISNAT(s(x)) → ISNAT(x)

The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

The set Q consists of the following terms:

f(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

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:

ISNAT(s(x)) → ISNAT(x)

R is empty.
The set Q consists of the following terms:

f(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

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(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ISNAT(s(x)) → ISNAT(x)

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:

  • ISNAT(s(x)) → ISNAT(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

COND(tt, x, y) → F(s(x), s(y))
F(x, y) → COND(and(isNat(x), isNat(y)), x, y)

The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

The set Q consists of the following terms:

f(x0, x1)
cond(tt, x0, x1)
isNat(s(x0))
isNat(0)
and(tt, tt)
and(ff, x0)
and(x0, ff)

We have to consider all minimal (P,Q,R)-chains.

(15) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

COND(tt, x, y) → F(s(x), s(y))
F(x, y) → COND(and(isNat(x), isNat(y)), x, y)

The TRS R consists of the following rules:

f(x, y) → cond(and(isNat(x), isNat(y)), x, y)
cond(tt, x, y) → f(s(x), s(y))
isNat(s(x)) → isNat(x)
isNat(0) → tt
and(tt, tt) → tt
and(ff, x) → ff
and(x, ff) → ff

Q is empty.
We have to consider all (P,Q,R)-chains.

(17) 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
COND(tt, zr0, zr2)[zr0 / s(zr0), zr2 / s(zr2)]n[zr0 / 0, zr2 / 0] → COND(tt, s(zr0), s(zr2))[zr0 / s(zr0), zr2 / s(zr2)]n[zr0 / 0, zr2 / 0]
This rule is correct for the QDP as the following derivation shows:

COND(tt, zr0, zr2)[zr0 / s(zr0), zr2 / s(zr2)]n[zr0 / 0, zr2 / 0] → COND(tt, s(zr0), s(zr2))[zr0 / s(zr0), zr2 / s(zr2)]n[zr0 / 0, zr2 / 0]
    by Equivalence by Domain Renaming of the lhs with [zl0 / zr0, zl2 / zr2]
    intermediate steps: Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
    COND(tt, zl1, zl3)[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 0] → COND(tt, s(zr1), s(zr3))[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 0]
        by Narrowing at position: [0]
        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
        COND(tt, zl1, zl3)[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 0] → COND(and(tt, tt), s(zr1), s(zr3))[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 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) - Equiv IPS (rhs) - Equiv IPS (lhs)
            COND(tt, zl1, zl3)[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / x1, zl1 / 0, zl3 / x1] → COND(and(tt, isNat(x1)), s(zr1), s(zr3))[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]n[zr1 / 0, zr3 / x1, zl1 / 0, zl3 / x1]
                by Narrowing at position: [0,0]
                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) - Equiv IPS (rhs) - Equiv IPS (lhs)
                COND(tt, zl1, x0)[zr1 / s(zr1), zl1 / s(zl1), x0 / s(x0)]n[zr1 / x1, zl1 / x1, x0 / y0] → COND(and(isNat(x1), isNat(y0)), s(zr1), s(x0))[zr1 / s(zr1), zl1 / s(zl1), x0 / s(x0)]n[zr1 / x1, zl1 / x1, x0 / y0]
                    by Narrowing at position: [0,1]
                    intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
                    COND(tt, x0, x1)[x0 / s(x0)]n[x0 / y0] → COND(and(isNat(y0), isNat(s(x1))), s(x0), s(x1))[x0 / s(x0)]n[x0 / y0]
                        by Narrowing at position: [0,0]
                        intermediate steps: Instantiate mu - Instantiate Sigma
                        COND(tt, x0, x1)[ ]n[ ] → COND(and(isNat(s(x0)), isNat(s(x1))), s(x0), s(x1))[ ]n[ ]
                            by Narrowing at position: []
                            intermediate steps: Instantiation
                            COND(tt, x, y)[ ]n[ ] → F(s(x), s(y))[ ]n[ ]
                                by Rule from TRS P

                            intermediate steps: Instantiation - Instantiation - Instantiation - Instantiation
                            F(x, y)[ ]n[ ] → COND(and(isNat(x), isNat(y)), x, y)[ ]n[ ]
                                by Rule from TRS P

                        intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Equiv DR (lhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
                        isNat(s(x))[x / s(x)]n[ ] → isNat(x)[ ]n[ ]
                            by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x)]
                            isNat(s(x))[ ]n[ ] → isNat(x)[ ]n[ ]
                                by Rule from TRS R

                    intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (lhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
                    isNat(s(x))[x / s(x)]n[ ] → isNat(x)[ ]n[ ]
                        by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x)]
                        isNat(s(x))[ ]n[ ] → isNat(x)[ ]n[ ]
                            by Rule from TRS R

                intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
                isNat(0)[ ]n[ ] → tt[ ]n[ ]
                    by Rule from TRS R

            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
            isNat(0)[ ]n[ ] → tt[ ]n[ ]
                by Rule from TRS R

        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
        and(tt, tt)[ ]n[ ] → tt[ ]n[ ]
            by Rule from TRS R

(18) NO