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:

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

(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(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(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
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)) = [1] + 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(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(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
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)) = x₁²   
POL(MINUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = [2]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(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(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
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)) = x₁²   
POL(MINUS(x₁, x₂)) = [1] + x₂   
POL(QUOT(x₁, 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   

(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

(16) BOUNDS(1, 1)