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

(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 CpxTRS 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:

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

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

Removed 1 trailing tuple parts

(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:

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

(5) CdtPolyRedPairProof (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:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
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)) = [1] + x₁   

(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))
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:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

(7) CdtPolyRedPairProof (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:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
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)) = [2]x₁ + [2]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)) = [1] + x₁   

(8) 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:

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:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

(9) CdtPolyRedPairProof (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:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
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)) = x₁²   
POL(MINUS(x₁, x₂)) = [2] + x₂   
POL(QUOT(x₁, x₂)) = [3] + x₂ + 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   

(10) 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: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:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))