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

(0) Obligation:

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


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
LOG(s(0)) → c4
QUOT(0, s(z0)) → c2

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

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)) = 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(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:

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

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 + x₁²   
POL(MIN(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = [2]x₁ + x₂²   
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)) = [2] + 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:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))

K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
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) CdtRuleRemovalProof (UPPER BOUND(ADD(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:

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

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)) = [1] + x₂   
POL(QUOT(x₁, x₂)) = [2]x₂ + [2]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   

(12) 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))))
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
MIN(s(z0), s(z1)) → c1(MIN(z0, z1))

Defined Rule Symbols:

min, quot

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(14) BOUNDS(1, 1)