### (0) Obligation:

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

pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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:

pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(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:

PRED(s(z0)) → c
MINUS(z0, 0) → c1
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c3
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(0)) → c5
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

PRED(s(z0)) → c
MINUS(z0, 0) → c1
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c3
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(0)) → c5
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

PRED, MINUS, QUOT, LOG

Compound Symbols:

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

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

Removed 4 trailing nodes:

PRED(s(z0)) → c
MINUS(z0, 0) → c1
LOG(s(0)) → c5
QUOT(0, s(z0)) → c3

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(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:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6

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

Removed 1 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(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:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

### (7) 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))))))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, pred, quot

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

### (9) 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))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
We considered the (Usable) Rules:

minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
pred(s(z0)) → z0
And the Tuples:

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

POL(0) = 0
POL(LOG(x1)) = x1
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [2] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
Defined Rule Symbols:

minus, pred, quot

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

### (11) 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)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
pred(s(z0)) → z0
And the Tuples:

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

POL(0) = 0
POL(LOG(x1)) = [2]x12
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = x1
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, pred, quot

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

### (13) 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.

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

minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
pred(s(z0)) → z0
And the Tuples:

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

POL(0) = [2]
POL(LOG(x1)) = [2]x12
POL(MINUS(x1, x2)) = [1] + [2]x2
POL(QUOT(x1, x2)) = [1] + x1·x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [2] + x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:none
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
Defined Rule Symbols:

minus, pred, quot

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

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

The set S is empty