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

(0) Obligation:

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


The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))

Rewrite Strategy: FULL

(1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

(2) Obligation:

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


The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))

Tuples:

QUOT(0, s(z0), s(z1)) → c
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

S tuples:

QUOT(0, s(z0), s(z1)) → c
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

K tuples:none
Defined Rule Symbols:

quot

Defined Pair Symbols:

QUOT

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

QUOT(0, s(z0), s(z1)) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))

Tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

S tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

K tuples:none
Defined Rule Symbols:

quot

Defined Pair Symbols:

QUOT

Compound Symbols:

c1, c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

S tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

QUOT

Compound Symbols:

c1, 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.

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))

We considered the (Usable) Rules:none
And the Tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(QUOT(x₁, x₂, x₃)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

S tuples:

QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))

K tuples:

QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))

Defined Rule Symbols:none

Defined Pair Symbols:

QUOT

Compound Symbols:

c1, c2

(11) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))

Now S is empty

(12) BOUNDS(1, 1)