Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

(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:

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(0, y, z) → MAX(y, z)
F(x, 0, z) → MAX(x, z)
F(x, y, 0) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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:

MAX(s(x), s(y)) → MAX(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

(11) Obligation:

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

MAX(s(x), s(y)) → MAX(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:

• MAX(s(x), s(y)) → MAX(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(15) 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.

(16) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(17) 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].

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

(18) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

(19) 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:

• MIN(s(x), s(y)) → MIN(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) Obligation:

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(22) 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.

(23) Obligation:

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(24) 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(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

(25) Obligation:

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(26) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z)))) at position [0,1] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(max(s(x), s(max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

(27) Obligation:

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

F(s(x), s(y), s(z)) → F(max(s(x), s(max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(28) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(max(s(x), s(max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z)))) at position [0] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

(29) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(30) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z)))) at position [1,0,1] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z))))

(31) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(32) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z)))) at position [1,0] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z))))

(33) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(34) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z)))) at position [1] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

(35) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
p(s(x)) → x
min(0, y) → 0
max(0, y) → y

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(36) 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.

(37) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
max(0, y) → y
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))

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

(38) 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].

p(s(x0))

(39) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
max(0, y) → y
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))

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

(40) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z)))) at position [2,1] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z))))

(41) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
max(0, y) → y
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))

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

(42) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z)))) at position [2] we obtained the following new rules [LPAR04]:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

(43) Obligation:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
max(0, y) → y
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

POL(0) = 0   
POL(F(x₁, x₂, x₃)) = x₂ + x₃   
POL(max(x₁, x₂)) = max(x₁, x₂)   
POL(min(x₁, x₂)) = min(x₁, x₂)   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

max(s(x), s(y)) → s(max(x, y))
max(0, y) → y
max(x, 0) → x
min(s(x), s(y)) → s(min(x, y))
min(x, 0) → 0
min(0, y) → 0

(45) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

max(s(x), s(y)) → s(max(x, y))
max(x, 0) → x
max(0, y) → y
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
min(0, y) → 0

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))

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

(46) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(47) TRUE