### (0) Obligation:

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

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

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

### (2) Obligation:

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

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(z0))) → prime1(s(s(z0)), s(z0))
prime1(z0, 0) → false
prime1(z0, s(0)) → true
prime1(z0, s(s(z1))) → and(not(divp(s(s(z1)), z0)), prime1(z0, s(z1)))
divp(z0, z1) → =(rem(z0, z1), 0)
Tuples:

PRIME(0) → c
PRIME(s(0)) → c1
PRIME(s(s(z0))) → c2(PRIME1(s(s(z0)), s(z0)))
PRIME1(z0, 0) → c3
PRIME1(z0, s(0)) → c4
PRIME1(z0, s(s(z1))) → c5(DIVP(s(s(z1)), z0), PRIME1(z0, s(z1)))
DIVP(z0, z1) → c6
S tuples:

PRIME(0) → c
PRIME(s(0)) → c1
PRIME(s(s(z0))) → c2(PRIME1(s(s(z0)), s(z0)))
PRIME1(z0, 0) → c3
PRIME1(z0, s(0)) → c4
PRIME1(z0, s(s(z1))) → c5(DIVP(s(s(z1)), z0), PRIME1(z0, s(z1)))
DIVP(z0, z1) → c6
K tuples:none
Defined Rule Symbols:

prime, prime1, divp

Defined Pair Symbols:

PRIME, PRIME1, DIVP

Compound Symbols:

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

### (5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

PRIME(s(s(z0))) → c2(PRIME1(s(s(z0)), s(z0)))
Removed 5 trailing nodes:

PRIME(0) → c
DIVP(z0, z1) → c6
PRIME1(z0, 0) → c3
PRIME1(z0, s(0)) → c4
PRIME(s(0)) → c1

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(z0))) → prime1(s(s(z0)), s(z0))
prime1(z0, 0) → false
prime1(z0, s(0)) → true
prime1(z0, s(s(z1))) → and(not(divp(s(s(z1)), z0)), prime1(z0, s(z1)))
divp(z0, z1) → =(rem(z0, z1), 0)
Tuples:

PRIME1(z0, s(s(z1))) → c5(DIVP(s(s(z1)), z0), PRIME1(z0, s(z1)))
S tuples:

PRIME1(z0, s(s(z1))) → c5(DIVP(s(s(z1)), z0), PRIME1(z0, s(z1)))
K tuples:none
Defined Rule Symbols:

prime, prime1, divp

Defined Pair Symbols:

PRIME1

Compound Symbols:

c5

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

Removed 1 trailing tuple parts

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(z0))) → prime1(s(s(z0)), s(z0))
prime1(z0, 0) → false
prime1(z0, s(0)) → true
prime1(z0, s(s(z1))) → and(not(divp(s(s(z1)), z0)), prime1(z0, s(z1)))
divp(z0, z1) → =(rem(z0, z1), 0)
Tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
S tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
K tuples:none
Defined Rule Symbols:

prime, prime1, divp

Defined Pair Symbols:

PRIME1

Compound Symbols:

c5

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

prime(0) → false
prime(s(0)) → false
prime(s(s(z0))) → prime1(s(s(z0)), s(z0))
prime1(z0, 0) → false
prime1(z0, s(0)) → true
prime1(z0, s(s(z1))) → and(not(divp(s(s(z1)), z0)), prime1(z0, s(z1)))
divp(z0, z1) → =(rem(z0, z1), 0)

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
S tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

PRIME1

Compound Symbols:

c5

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

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

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
We considered the (Usable) Rules:none
And the Tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(PRIME1(x1, x2)) = x2
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
S tuples:none
K tuples:

PRIME1(z0, s(s(z1))) → c5(PRIME1(z0, s(z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

PRIME1

Compound Symbols:

c5

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

The set S is empty