### (0) Obligation:

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

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Rewrite Strategy: FULL

### (1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)

The following defined symbols can occur below the 0th argument of plus: plus, times

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

### (2) Obligation:

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

The TRS R consists of the following rules:

times(x, s(y)) → plus(times(x, y), x)
plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
times(x, 0) → 0

Rewrite Strategy: FULL

### (3) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
times([], s(y))

The defined contexts are:
plus([], x1)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (4) 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:

times(x, s(y)) → plus(times(x, y), x)
plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
times(x, 0) → 0

Rewrite Strategy: INNERMOST

### (5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, s(z1)) → plus(times(z0, z1), z0)
times(z0, 0) → 0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(z0, 0) → z0
Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
TIMES(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
PLUS(z0, 0) → c3
S tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
TIMES(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
PLUS(z0, 0) → c3
K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, c1, c2, c3

### (7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

PLUS(z0, 0) → c3
TIMES(z0, 0) → c1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, s(z1)) → plus(times(z0, z1), z0)
times(z0, 0) → 0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(z0, 0) → z0
Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, 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.

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(PLUS(x1, x2)) = 0
POL(TIMES(x1, x2)) = x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(plus(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = 0

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, s(z1)) → plus(times(z0, z1), z0)
times(z0, 0) → 0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(z0, 0) → z0
Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, 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.

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(PLUS(x1, x2)) = [1] + x2
POL(TIMES(x1, x2)) = [2]x2 + x1·x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(plus(x1, x2)) = [2]x1 + [2]x2 + [2]x22 + x1·x2 + [2]x12
POL(s(x1)) = [2] + x1
POL(times(x1, x2)) = x1 + [2]x1·x2 + [2]x12

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, s(z1)) → plus(times(z0, z1), z0)
times(z0, 0) → 0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(z0, 0) → z0
Tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:none
K tuples:

TIMES(z0, s(z1)) → c(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, c2

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

The set S is empty