(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
The TRS R 2 is
h(true, x, y) → h(gt(x, y), s(x), s(y))
The signature Sigma is {
h}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
The set Q consists of the following terms:
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:
H(true, x, y) → H(gt(x, y), s(x), s(y))
H(true, x, y) → GT(x, y)
GT(s(x), s(y)) → GT(x, y)
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
The set Q consists of the following terms:
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:
GT(s(x), s(y)) → GT(x, y)
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
The set Q consists of the following terms:
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:
GT(s(x), s(y)) → GT(x, y)
R is empty.
The set Q consists of the following terms:
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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].
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(x), s(y)) → GT(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:
- GT(s(x), s(y)) → GT(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:
H(true, x, y) → H(gt(x, y), s(x), s(y))
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
The set Q consists of the following terms:
h(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:
H(true, x, y) → H(gt(x, y), s(x), s(y))
The TRS R consists of the following rules:
h(true, x, y) → h(gt(x, y), s(x), s(y))
gt(0, x) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(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
H(true, s(s(zr0)), s(zr1))[zr0 / s(zr0), zr1 / s(zr1)]n[zr1 / 0] → H(true, s(s(s(zr0))), s(s(zr1)))[zr0 / s(zr0), zr1 / s(zr1)]n[zr1 / 0]
This rule is correct for the QDP as the following derivation shows:
H(true, s(s(zr0)), s(zr1))[zr0 / s(zr0), zr1 / s(zr1)]n[zr1 / 0] → H(true, s(s(s(zr0))), s(s(zr1)))[zr0 / s(zr0), zr1 / s(zr1)]n[zr1 / 0]
by Equivalence by Irrelevant Pattern Substitutions σ: [zr0 / s(zr0), zr1 / s(zr1)] µ: [zr1 / 0]
intermediate steps: Equiv IPS (lhs) - Instantiate mu - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
H(true, s(s(zl2)), s(zl3))[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / y0, zr3 / 0, zl2 / y0, zl3 / 0] → H(true, s(s(s(zr2))), s(s(zr3)))[zr2 / s(zr2), zr3 / s(zr3), zl2 / s(zl2), zl3 / s(zl3)]n[zr2 / y0, zr3 / 0, zl2 / y0, zl3 / 0]
by Narrowing at position: [0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv Sµ (rhs) - Equiv Sµ (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
H(true, s(zs2), s(zs3))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0] → H(gt(y1, y0), s(s(zs2)), s(s(zs3)))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation - Instantiation
H(true, x, y)[ ]n[ ] → H(gt(x, y), s(x), s(y))[ ]n[ ]
by Rule from TRS P
intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
gt(s(x), s(y))[x / s(x), y / s(y)]n[ ] → gt(x, y)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
gt(s(x), s(y))[ ]n[ ] → gt(x, y)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation
gt(s(x), 0)[ ]n[ ] → true[ ]n[ ]
by Rule from TRS R
(18) NO