### (0) Obligation:

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(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:

min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(z0, 0) → c
MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(0)) → c4
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

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

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

MIN(z0, 0) → c
QUOT(0, s(z0)) → c2
LOG(s(0)) → c4

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

### (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:

min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

min, quot

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

### (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))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
We considered the (Usable) Rules:

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

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(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(LOG(x1)) = [2]x1
POL(MIN(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1, x2)) = x1 + x2
POL(min(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
Defined Rule Symbols:

min, quot

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

### (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.

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
We considered the (Usable) Rules:

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

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(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(LOG(x1)) = x12
POL(MIN(x1, x2)) = x2
POL(QUOT(x1, x2)) = [3] + x1·x2
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1, x2)) = x1 + x2
POL(min(x1, x2)) = x1
POL(quot(x1, x2)) = x1
POL(s(x1)) = [1] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:none
K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
Defined Rule Symbols:

min, quot

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

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

The set S is empty