```
We consider the following Problem:

Strict Trs:
{  q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))
, q(s(x), s(y), z) -> q(x, y, z)
, q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
We consider the following Problem:

Strict Trs:
{  q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))
, q(s(x), s(y), z) -> q(x, y, z)
, q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
The weightgap principle applies, where following rules are oriented strictly:

TRS Component: {q(0(), s(y), s(z)) -> 0()}

Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(q) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
0() = 

s(x1) = [1 2] x1 + 
[0 0]      
q(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + 
[0 0]      [0 0]      [0 0]      

The strictly oriented rules are moved into the weak component.

We consider the following Problem:

Strict Trs:
{  q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))
, q(s(x), s(y), z) -> q(x, y, z)}
Weak Trs: {q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
The weightgap principle applies, where following rules are oriented strictly:

TRS Component: {q(s(x), s(y), z) -> q(x, y, z)}

Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(q) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
0() = 

s(x1) = [1 3] x1 + 
[0 0]      
q(x1, x2, x3) = [1 1] x1 + [0 0] x2 + [0 0] x3 + 
[0 0]      [0 0]      [0 1]      

The strictly oriented rules are moved into the weak component.

We consider the following Problem:

Strict Trs: {q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))}
Weak Trs:
{  q(s(x), s(y), z) -> q(x, y, z)
, q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
We consider the following Problem:

Strict Trs: {q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))}
Weak Trs:
{  q(s(x), s(y), z) -> q(x, y, z)
, q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
We have computed the following dependency pairs

Strict DPs: {q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(0(), s(y), s(z)) -> c_3()}

We consider the following Problem:

Strict DPs: {q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
Strict Trs: {q(x, 0(), s(z)) -> s(q(x, s(z), s(z)))}
Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(0(), s(y), s(z)) -> c_3()}
Weak Trs:
{  q(s(x), s(y), z) -> q(x, y, z)
, q(0(), s(y), s(z)) -> 0()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
No rule is usable.

We consider the following Problem:

Strict DPs: {q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(0(), s(y), s(z)) -> c_3()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
We consider the following Problem:

Strict DPs: {q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(0(), s(y), s(z)) -> c_3()}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
We use following congruence DG for path analysis

->1:{1,2}                                                   [   YES(O(1),O(1))   ]
|
`->2:{3}                                                 [   YES(O(1),O(1))   ]

Here dependency-pairs are as follows:

Strict DPs:
{1: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
WeakDPs DPs:
{  2: q^#(s(x), s(y), z) -> q^#(x, y, z)
, 3: q^#(0(), s(y), s(z)) -> c_3()}

* Path 1:{1,2}: YES(O(1),O(1))
----------------------------

We consider the following Problem:

Strict DPs: {q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the the dependency-graph

1: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))

together with the congruence-graph

->1:{1}                                                     Noncyclic, trivial, SCC

Here dependency-pairs are as follows:

Strict DPs:
{1: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

The following rules are either leafs or part of trailing weak paths, and thus they can be removed:

{1: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
Empty rules are trivially bounded

* Path 1:{1,2}->2:{3}: YES(O(1),O(1))
-----------------------------------

We consider the following Problem:

Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the the dependency-graph

1: q^#(s(x), s(y), z) -> q^#(x, y, z)
-->_1 q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z)) :2
-->_1 q^#(s(x), s(y), z) -> q^#(x, y, z) :1

2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))
-->_1 q^#(s(x), s(y), z) -> q^#(x, y, z) :1

together with the congruence-graph

->1:{1,2}                                                   Weak SCC

Here dependency-pairs are as follows:

WeakDPs DPs:
{  1: q^#(s(x), s(y), z) -> q^#(x, y, z)
, 2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

The following rules are either leafs or part of trailing weak paths, and thus they can be removed:

{  1: q^#(s(x), s(y), z) -> q^#(x, y, z)
, 2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
Empty rules are trivially bounded

* Path 1:{1,2}->2:{3}: YES(O(1),O(1))
-----------------------------------

We consider the following Problem:

Weak DPs:
{  q^#(s(x), s(y), z) -> q^#(x, y, z)
, q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}
StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the the dependency-graph

1: q^#(s(x), s(y), z) -> q^#(x, y, z)
-->_1 q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z)) :2
-->_1 q^#(s(x), s(y), z) -> q^#(x, y, z) :1

2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))
-->_1 q^#(s(x), s(y), z) -> q^#(x, y, z) :1

together with the congruence-graph

->1:{1,2}                                                   Weak SCC

Here dependency-pairs are as follows:

WeakDPs DPs:
{  1: q^#(s(x), s(y), z) -> q^#(x, y, z)
, 2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

The following rules are either leafs or part of trailing weak paths, and thus they can be removed:

{  1: q^#(s(x), s(y), z) -> q^#(x, y, z)
, 2: q^#(x, 0(), s(z)) -> q^#(x, s(z), s(z))}

We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
We consider the following Problem:

StartTerms: basic terms
Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
Empty rules are trivially bounded