### (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:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Rewrite Strategy: FULL

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

The following defined symbols can occur below the 0th argument of h: d
The following defined symbols can occur below the 0th argument of d: d

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))

### (2) 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:

h(e(x), y) → h(d(x, y), s(y))
g(e(x), e(y)) → e(g(x, y))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

### (4) 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:

h(e(x), y) → h(d(x, y), s(y))
g(e(x), e(y)) → e(g(x, y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(e(z0), z1) → h(d(z0, z1), s(z1))
g(e(z0), e(z1)) → e(g(z0, z1))
Tuples:

H(e(z0), z1) → c(H(d(z0, z1), s(z1)))
G(e(z0), e(z1)) → c1(G(z0, z1))
S tuples:

H(e(z0), z1) → c(H(d(z0, z1), s(z1)))
G(e(z0), e(z1)) → c1(G(z0, z1))
K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

H(e(z0), z1) → c(H(d(z0, z1), s(z1)))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(e(z0), z1) → h(d(z0, z1), s(z1))
g(e(z0), e(z1)) → e(g(z0, z1))
Tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
S tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

G

Compound Symbols:

c1

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(e(z0), z1) → h(d(z0, z1), s(z1))
g(e(z0), e(z1)) → e(g(z0, z1))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
S tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

### (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.

G(e(z0), e(z1)) → c1(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1, x2)) = [2]x22
POL(c1(x1)) = x1
POL(e(x1)) = [1] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
S tuples:none
K tuples:

G(e(z0), e(z1)) → c1(G(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

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

The set S is empty