### (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:

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: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
minus([], s(y))
minus(x, s([]))

The defined contexts are:
if([], 0, x1)
if(x0, 0, [])
p([])
minus(x0, [])
le(x0, s([]))
p(s([]))
le(x0, [])

[] just represents basic- or constructor-terms in the following defined contexts:
if([], 0, x1)
minus(x0, [])

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:

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

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

Converted Cpx (relative) TRS to CDT

### (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:

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

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

Removed 7 trailing nodes:

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

### (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(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

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

Removed 3 trailing tuple parts

### (8) 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

### (9) 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

### (10) 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

### (11) 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))

### (12) 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

### (13) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

p(s(z0)) → z0

### (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))
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

### (15) 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)) = 0
POL(MINUS(x1, x2)) = 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:

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

### (17) 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)) = x1
POL(MINUS(x1, x2)) = x1·x2
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(s(x1)) = [2] + x1

### (18) 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

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

The set S is empty