### (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(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
f(x, c([]))

The defined contexts are:
f(x0, s([]))
f(x0, s(c([])))

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(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))
S tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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

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

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x2
POL(c(x1)) = [1] + x1
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(f(x1, x2)) = x2 + x22 + x1·x2
POL(s(x1)) = 0

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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

Use instantiation to replace F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) by

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

### (9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

### (11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

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

F(s(z0), z1) → c2(F(z0, s(c(z1))))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:none
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

The set S is empty