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:

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 CpxTRS 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(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

(3) 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))) → c5(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(min(z0, z1), s(z1)))
min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, 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   

(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))

K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))

Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

(5) 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.

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))

We considered the (Usable) Rules:

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

(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)))
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:

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))))
MIN(s(z0), s(z1)) → c1(MIN(z0, z1))

Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

(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)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))

We considered the (Usable) Rules:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1)))
min(z0, 0) → z0
min(s(z0), s(z1)) → min(z0, 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]x₁²   
POL(MIN(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = [2]x₁   
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)))
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: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, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))