(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, y) → cond(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
The TRS R 2 is
f(x, y) → cond(lt(x, 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(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
f(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(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(x, y) → COND(lt(x, y), x, y)
F(x, y) → LT(x, y)
COND(tt, x, y) → F(s(x), s(y))
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
f(x, y) → cond(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
f(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(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 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
f(x, y) → cond(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
f(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(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:
LT(s(x), s(y)) → LT(x, y)
R is empty.
The set Q consists of the following terms:
f(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(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(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(s(x0), s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(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:
- LT(s(x), s(y)) → LT(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:
COND(tt, x, y) → F(s(x), s(y))
F(x, y) → COND(lt(x, y), x, y)
The TRS R consists of the following rules:
f(x, y) → cond(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
f(x0, x1)
cond(tt, x0, x1)
lt(0, x0)
lt(s(x0), s(x1))
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(lt(x, y), x, y)
The TRS R consists of the following rules:
f(x, y) → cond(lt(x, y), x, y)
cond(tt, x, y) → f(s(x), s(y))
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
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, zr1)[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0] → COND(tt, s(zr0), s(zr1))[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:
COND(tt, zr0, zr1)[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0] → COND(tt, s(zr0), s(zr1))[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0]
by Equivalence by Irrelevant Pattern Substitutions σ: [zr0 / s(zr0), zr1 / s(zr1)] µ: [zr0 / 0]
intermediate steps: Equiv IPS (lhs) - Instantiate mu - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
COND(tt, zl2, zl3)[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / 0, zr3 / x1, zl2 / 0, zl3 / x1] → COND(tt, s(zr2), s(zr3))[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / 0, zr3 / x1, zl2 / 0, zl3 / x1]
by Narrowing at position: [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)
COND(tt, x0, x1)[x0 / s(x0), x1 / s(x1)]n[x0 / y1, x1 / y0] → COND(lt(y1, y0), s(x0), s(x1))[x0 / s(x0), x1 / s(x1)]n[x0 / y1, x1 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma
COND(tt, x0, x1)[ ]n[ ] → COND(lt(s(x0), 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(lt(x, 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) - Instantiation - Equiv DR (lhs)
lt(s(x), s(y))[x / s(x), y / s(y)]n[ ] → lt(x, y)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
lt(s(x), s(y))[ ]n[ ] → lt(x, y)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Instantiation
lt(0, y)[ ]n[ ] → tt[ ]n[ ]
by Rule from TRS R
(18) NO