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:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
admit([], .(u, .(v, .(w, z))))
admit(x, .([], .(v, .(w, z))))
admit(x, .(u, .([], .(w, z))))

The defined contexts are:
cond(=(sum(x0, x1, x2), w), .(x3, .(x4, .(w, []))))

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, 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:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0

Tuples:

ADMIT(z0, nil) → c
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
COND(true, z0) → c2

S tuples:

ADMIT(z0, nil) → c
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
COND(true, z0) → c2

K tuples:none
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT, COND

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

ADMIT(z0, nil) → c
COND(true, z0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0

Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))

S tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))

K tuples:none
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0

Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

S tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

K tuples:none
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

S tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

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

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

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

POL(.(x₁, x₂)) = [2] + x₁ + x₂   
POL(ADMIT(x₁, x₂)) = x₁ + [2]x₂   
POL(c1(x₁)) = x₁   
POL(carry(x₁, x₂, x₃)) = [3] + x₁ + x₃   
POL(w) = [2]   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

S tuples:none
K tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

Defined Rule Symbols:none

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

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

The set S is empty

(14) BOUNDS(1, 1)