### (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:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
f(s([]))
g(s([]), y)

The defined contexts are:
g(x0, [])
g(x0, s([]))
+([], 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:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

F(0) → c
F(s(z0)) → c1(G(z0, s(z0)))
G(0, z0) → c2
G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

F(0) → c
F(s(z0)) → c1(G(z0, s(z0)))
G(0, z0) → c2
G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

f, g, +

Defined Pair Symbols:

F, G, +'

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

### (5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

F(s(z0)) → c1(G(z0, s(z0)))
Removed 3 trailing nodes:

+'(z0, 0) → c5
F(0) → c
G(0, z0) → c2

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

f, g, +

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

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

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(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(G(x1, x2)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(s(x1)) = [1] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

### (11) 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)) → c6(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x1, x2)) = [2]x1 + [2]x22 + [2]x1·x2
POL(+'(x1, x2)) = [2]x2
POL(0) = [2]
POL(G(x1, x2)) = [2]x1 + x12
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(s(x1)) = [1] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:none
K tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))
Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

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

The set S is empty