(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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
Tuples:

D'(t) → c
D'(constant) → c1
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:

D'(t) → c
D'(constant) → c1
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 2 trailing nodes:

D'(constant) → c1
D'(t) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

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

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
We considered the (Usable) Rules:none
And the Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(*(x1, x2)) = [1] + x1 + x2
POL(+(x1, x2)) = x1 + x2
POL(-(x1, x2)) = [1] + x1 + x2
POL(D'(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
K tuples:

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
We considered the (Usable) Rules:none
And the Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(*(x1, x2)) = [2] + x1 + x2
POL(+(x1, x2)) = [3] + x1 + x2
POL(-(x1, x2)) = [2] + x1 + x2
POL(D'(x1)) = [1] + [3]x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:none
K tuples:

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

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

The set S is empty