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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtUsableRulesProof (⇔, 0 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 2 ms)
↳14 CdtProblem
↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 15 ms)
↳16 CdtProblem
↳17 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳18 BOUNDS(1, 1)

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(z0, 0) → z0
minus(z0, s(z1)) → if(le(z0, s(z1)), 0, p(minus(z0, p(s(z1)))))
if(true, z0, z1) → z0
if(false, z0, z1) → z1
Tuples:

P(0) → c
P(s(z0)) → c1
LE(0, z0) → c2
LE(s(z0), 0) → c3
LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, 0) → c5
MINUS(z0, s(z1)) → c6(IF(le(z0, s(z1)), 0, p(minus(z0, p(s(z1))))), LE(z0, s(z1)), P(minus(z0, p(s(z1)))), MINUS(z0, p(s(z1))), P(s(z1)))
IF(true, z0, z1) → c7
IF(false, z0, z1) → c8
S tuples:

P(0) → c
P(s(z0)) → c1
LE(0, z0) → c2
LE(s(z0), 0) → c3
LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, 0) → c5
MINUS(z0, s(z1)) → c6(IF(le(z0, s(z1)), 0, p(minus(z0, p(s(z1))))), LE(z0, s(z1)), P(minus(z0, p(s(z1)))), MINUS(z0, p(s(z1))), P(s(z1)))
IF(true, z0, z1) → c7
IF(false, z0, z1) → c8
K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

P, LE, MINUS, IF

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 7 trailing nodes:

IF(true, z0, z1) → c7
IF(false, z0, z1) → c8
LE(0, z0) → c2
P(s(z0)) → c1
P(0) → c
MINUS(z0, 0) → c5
LE(s(z0), 0) → c3

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(z0, 0) → z0
minus(z0, s(z1)) → if(le(z0, s(z1)), 0, p(minus(z0, p(s(z1)))))
if(true, z0, z1) → z0
if(false, z0, z1) → z1
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(IF(le(z0, s(z1)), 0, p(minus(z0, p(s(z1))))), LE(z0, s(z1)), P(minus(z0, p(s(z1)))), MINUS(z0, p(s(z1))), P(s(z1)))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(IF(le(z0, s(z1)), 0, p(minus(z0, p(s(z1))))), LE(z0, s(z1)), P(minus(z0, p(s(z1)))), MINUS(z0, p(s(z1))), P(s(z1)))
K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(z0, 0) → z0
minus(z0, s(z1)) → if(le(z0, s(z1)), 0, p(minus(z0, p(s(z1)))))
if(true, z0, z1) → z0
if(false, z0, z1) → z1
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(LE(z0, s(z1)), MINUS(z0, p(s(z1))))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(LE(z0, s(z1)), MINUS(z0, p(s(z1))))
K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

p(0) → 0
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(z0, 0) → z0
minus(z0, s(z1)) → if(le(z0, s(z1)), 0, p(minus(z0, p(s(z1)))))
if(true, z0, z1) → z0
if(false, z0, z1) → z1

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(LE(z0, s(z1)), MINUS(z0, p(s(z1))))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, s(z1)) → c6(LE(z0, s(z1)), MINUS(z0, p(s(z1))))
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace MINUS(z0, s(z1)) → c6(LE(z0, s(z1)), MINUS(z0, p(s(z1)))) by

MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

p(s(z0)) → z0

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(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.

MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
We considered the (Usable) Rules:none
And the Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(LE(x1, x2)) = [3]   
POL(MINUS(x1, x2)) = x2   
POL(c4(x1)) = x1   
POL(c6(x1, x2)) = x1 + x2   
POL(s(x1)) = [4] + x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
K tuples:

MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
Defined Rule Symbols:none

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

LE(s(z0), s(z1)) → c4(LE(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(LE(x1, x2)) = [2] + x2   
POL(MINUS(x1, x2)) = x2 + [2]x22 + [2]x1·x2   
POL(c4(x1)) = x1   
POL(c6(x1, x2)) = x1 + x2   
POL(s(x1)) = [1] + x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
S tuples:none
K tuples:

MINUS(x0, s(z0)) → c6(LE(x0, s(z0)), MINUS(x0, z0))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

LE, MINUS

Compound Symbols:

c4, c6

(17) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(18) BOUNDS(1, 1)