### (0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Rewrite Strategy: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
sum(s([]))

The defined contexts are:
+([], s(x1))
+([], x1)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Rewrite Strategy: INNERMOST

### (3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SUM(0) → c
SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, 0) → c2
+'(z0, s(z1)) → c3(+'(z0, z1))
S tuples:

SUM(0) → c
SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, 0) → c2
+'(z0, s(z1)) → c3(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c, c1, c2, c3

### (5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

SUM(0) → c
+'(z0, 0) → c2

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
S tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

### (7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
We considered the (Usable) Rules:none
And the Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x1, x2)) = 0
POL(+'(x1, x2)) = 0
POL(0) = 0
POL(SUM(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(s(x1)) =  + x1
POL(sum(x1)) = 0

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
S tuples:

+'(z0, s(z1)) → c3(+'(z0, z1))
K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

### (9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

+'(z0, s(z1)) → c3(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x1, x2)) = x2 + x1·x2
POL(+'(x1, x2)) =  + x2
POL(0) = 
POL(SUM(x1)) = x1 + x12
POL(c1(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(s(x1)) =  + x1
POL(sum(x1)) = 

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
S tuples:none
K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

### (11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty