### (0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

circ(cons(a, s), t) → cons(msubst(a, t), circ(s, t))
circ(cons(lift, s), cons(a, t)) → cons(a, circ(s, t))
circ(cons(lift, s), cons(lift, t)) → cons(lift, circ(s, t))
circ(circ(s, t), u) → circ(s, circ(t, u))
circ(s, id) → s
circ(id, s) → s
circ(cons(lift, s), circ(cons(lift, t), u)) → circ(cons(lift, circ(s, t)), u)
subst(a, id) → a
msubst(a, id) → a
msubst(msubst(a, s), t) → msubst(a, circ(s, t))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2))
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2))
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1))
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2))
circ(z0, id) → z0
circ(id, z0) → z0
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2)
subst(z0, id) → z0
msubst(z0, id) → z0
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
CIRC(z0, id) → c4
CIRC(id, z0) → c5
CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1))
SUBST(z0, id) → c7
MSUBST(z0, id) → c8
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
CIRC(z0, id) → c4
CIRC(id, z0) → c5
CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1))
SUBST(z0, id) → c7
MSUBST(z0, id) → c8
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:none
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, SUBST, MSUBST

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1))
Removed 4 trailing nodes:

CIRC(id, z0) → c5
CIRC(z0, id) → c4
SUBST(z0, id) → c7
MSUBST(z0, id) → c8

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2))
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2))
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1))
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2))
circ(z0, id) → z0
circ(id, z0) → z0
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2)
subst(z0, id) → z0
msubst(z0, id) → z0
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:none
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2)
subst(z0, id) → z0

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2))
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2))
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1))
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2))
circ(z0, id) → z0
circ(id, z0) → z0
msubst(z0, id) → z0
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:none
Defined Rule Symbols:

circ, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

### (7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CIRC(x1, x2)) = [4]x1
POL(MSUBST(x1, x2)) = [2]x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(circ(x1, x2)) = [1] + [2]x1 + [3]x2
POL(cons(x1, x2)) = x1 + x2
POL(id) = 0
POL(lift) = 0
POL(msubst(x1, x2)) = [4] + [2]x1 + [2]x2

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2))
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2))
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1))
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2))
circ(z0, id) → z0
circ(id, z0) → z0
msubst(z0, id) → z0
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
K tuples:

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
Defined Rule Symbols:

circ, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

### (9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CIRC(x1, x2)) = [3] + x1
POL(MSUBST(x1, x2)) = [2] + x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(circ(x1, x2)) = [4] + x1 + x2
POL(cons(x1, x2)) = [5] + x1 + x2
POL(id) = 0
POL(lift) = 0
POL(msubst(x1, x2)) = [3] + x1 + x2

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2))
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2))
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1))
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2))
circ(z0, id) → z0
circ(id, z0) → z0
msubst(z0, id) → z0
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:none
K tuples:

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2))
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2))
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1))
Defined Rule Symbols:

circ, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

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

The set S is empty