### (0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(0)) → 0
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(0, z0) → c3
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(true, s(z0), z1) → c5
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(0, s(z0)) → c7
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(0)) → c9
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(0, z0) → c3
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(true, s(z0), z1) → c5
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(0, s(z0)) → c7
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(0)) → c9
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

le, minus, if_minus, quot, log

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10

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

Removed 6 trailing nodes:

LE(0, z0) → c
LE(s(z0), 0) → c1
IF_MINUS(true, s(z0), z1) → c5
MINUS(0, z0) → c3
LOG(s(0)) → c9
QUOT(0, s(z0)) → c7

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

le, minus, if_minus, quot, log

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

log(s(0)) → 0
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

le, minus, if_minus, quot

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
We considered the (Usable) Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
if_minus(true, s(z0), z1) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]
POL(IF_MINUS(x1, x2, x3)) = 0
POL(LE(x1, x2)) = 0
POL(LOG(x1)) = x1
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = [3]
POL(c10(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(if_minus(x1, x2, x3)) = x2
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [4] + x1
POL(true) = 0

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
Defined Rule Symbols:

le, minus, if_minus, quot

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
if_minus(true, s(z0), z1) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(IF_MINUS(x1, x2, x3)) = 0
POL(LE(x1, x2)) = 0
POL(LOG(x1)) = x12
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = x1
POL(c10(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(if_minus(x1, x2, x3)) = x2
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

le, minus, if_minus, quot

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (11) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

LE(s(z0), s(z1)) → c2(LE(z0, z1))
We considered the (Usable) Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
if_minus(true, s(z0), z1) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(IF_MINUS(x1, x2, x3)) = x2·x3
POL(LE(x1, x2)) = x2
POL(LOG(x1)) = x13
POL(MINUS(x1, x2)) = x2 + x1·x2
POL(QUOT(x1, x2)) = x12·x2
POL(c10(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(if_minus(x1, x2, x3)) = x2
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LE(s(z0), s(z1)) → c2(LE(z0, z1))
Defined Rule Symbols:

le, minus, if_minus, quot

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (13) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
We considered the (Usable) Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
if_minus(true, s(z0), z1) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(IF_MINUS(x1, x2, x3)) = x2
POL(LE(x1, x2)) = 0
POL(LOG(x1)) = x13
POL(MINUS(x1, x2)) = x1
POL(QUOT(x1, x2)) = [1] + x12 + x12·x2
POL(c10(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(if_minus(x1, x2, x3)) = x2
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → if_minus(le(s(z0), z1), s(z0), z1)
if_minus(true, s(z0), z1) → 0
if_minus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

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

LOG(s(s(z0))) → c10(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LE(s(z0), s(z1)) → c2(LE(z0, z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
Defined Rule Symbols:

le, minus, if_minus, quot

Defined Pair Symbols:

LE, MINUS, IF_MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6, c8, c10

### (15) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

MINUS(s(z0), z1) → c4(IF_MINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IF_MINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
LE(s(z0), s(z1)) → c2(LE(z0, z1))
Now S is empty