### (0) Obligation:

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

rec(rec(x)) → sent(rec(x))
rec(sent(x)) → sent(rec(x))
rec(no(x)) → sent(rec(x))
rec(bot) → up(sent(bot))
rec(up(x)) → up(rec(x))
sent(up(x)) → up(sent(x))
no(up(x)) → up(no(x))
top(rec(up(x))) → top(check(rec(x)))
top(sent(up(x))) → top(check(rec(x)))
top(no(up(x))) → top(check(rec(x)))
check(up(x)) → up(check(x))
check(sent(x)) → sent(check(x))
check(rec(x)) → rec(check(x))
check(no(x)) → no(check(x))
check(no(x)) → no(x)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
rec(up(z0)) → up(rec(z0))
sent(up(z0)) → up(sent(z0))
no(up(z0)) → up(no(z0))
top(rec(up(z0))) → top(check(rec(z0)))
top(sent(up(z0))) → top(check(rec(z0)))
top(no(up(z0))) → top(check(rec(z0)))
check(up(z0)) → up(check(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
Tuples:

REC(rec(z0)) → c(SENT(rec(z0)), REC(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(bot) → c3(SENT(bot))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
TOP(rec(up(z0))) → c7(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(sent(up(z0))) → c8(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(no(up(z0))) → c9(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(rec(z0)) → c12(REC(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
S tuples:

REC(rec(z0)) → c(SENT(rec(z0)), REC(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(bot) → c3(SENT(bot))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
TOP(rec(up(z0))) → c7(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(sent(up(z0))) → c8(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(no(up(z0))) → c9(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(rec(z0)) → c12(REC(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
K tuples:none
Defined Rule Symbols:

rec, sent, no, top, check

Defined Pair Symbols:

REC, SENT, NO, TOP, CHECK

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

TOP(rec(up(z0))) → c7(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(sent(up(z0))) → c8(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
TOP(no(up(z0))) → c9(TOP(check(rec(z0))), CHECK(rec(z0)), REC(z0))
Removed 1 trailing nodes:

REC(bot) → c3(SENT(bot))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
rec(up(z0)) → up(rec(z0))
sent(up(z0)) → up(sent(z0))
no(up(z0)) → up(no(z0))
top(rec(up(z0))) → top(check(rec(z0)))
top(sent(up(z0))) → top(check(rec(z0)))
top(no(up(z0))) → top(check(rec(z0)))
check(up(z0)) → up(check(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
Tuples:

REC(rec(z0)) → c(SENT(rec(z0)), REC(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(rec(z0)) → c12(REC(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
S tuples:

REC(rec(z0)) → c(SENT(rec(z0)), REC(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(rec(z0)) → c12(REC(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
K tuples:none
Defined Rule Symbols:

rec, sent, no, top, check

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c, c1, c2, c4, c5, c6, c10, c11, c12, c13, c14

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

Removed 2 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
rec(up(z0)) → up(rec(z0))
sent(up(z0)) → up(sent(z0))
no(up(z0)) → up(no(z0))
top(rec(up(z0))) → top(check(rec(z0)))
top(sent(up(z0))) → top(check(rec(z0)))
top(no(up(z0))) → top(check(rec(z0)))
check(up(z0)) → up(check(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
K tuples:none
Defined Rule Symbols:

rec, sent, no, top, check

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

rec(up(z0)) → up(rec(z0))
top(rec(up(z0))) → top(check(rec(z0)))
top(sent(up(z0))) → top(check(rec(z0)))
top(no(up(z0))) → top(check(rec(z0)))
check(up(z0)) → up(check(z0))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
K tuples:none
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (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.

CHECK(no(z0)) → c14(NO(z0))
We considered the (Usable) Rules:none
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [4]
POL(NO(x1)) = 0
POL(REC(x1)) = 0
POL(SENT(x1)) = 0
POL(bot) = [3]
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = x1
POL(no(x1)) = [1] + [3]x1
POL(rec(x1)) = [2]
POL(sent(x1)) = 0
POL(up(x1)) = [5]

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
K tuples:

CHECK(no(z0)) → c14(NO(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (11) 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.

CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
We considered the (Usable) Rules:none
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [3] + [4]x1
POL(NO(x1)) = [4]
POL(REC(x1)) = 0
POL(SENT(x1)) = 0
POL(bot) = [3]
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = [4]x1
POL(no(x1)) = [4] + [4]x1
POL(rec(x1)) = [2]x1
POL(sent(x1)) = [2]x1
POL(up(x1)) = [3] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
K tuples:

CHECK(no(z0)) → c14(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (13) 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.

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
We considered the (Usable) Rules:none
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [3] + [4]x1
POL(NO(x1)) = 0
POL(REC(x1)) = x1
POL(SENT(x1)) = 0
POL(bot) = [2]
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = [4] + [2]x1
POL(no(x1)) = x1
POL(rec(x1)) = [1] + x1
POL(sent(x1)) = [5] + x1
POL(up(x1)) = x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
K tuples:

CHECK(no(z0)) → c14(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (15) 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.

NO(up(z0)) → c6(NO(z0))
We considered the (Usable) Rules:

check(no(z0)) → no(check(z0))
no(up(z0)) → up(no(z0))
check(no(z0)) → no(z0)
check(rec(z0)) → rec(check(z0))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [2]x1
POL(NO(x1)) = [3] + x1
POL(REC(x1)) = 0
POL(SENT(x1)) = 0
POL(bot) = 0
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = [4]x1
POL(no(x1)) = [4] + [4]x1
POL(rec(x1)) = [2]x1
POL(sent(x1)) = [3] + [2]x1
POL(up(x1)) = [3] + x1

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
K tuples:

CHECK(no(z0)) → c14(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
NO(up(z0)) → c6(NO(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (17) 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.

REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
We considered the (Usable) Rules:none
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [5]
POL(NO(x1)) = 0
POL(REC(x1)) = [2]x1
POL(SENT(x1)) = 0
POL(bot) = [5]
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = x1
POL(no(x1)) = [4] + [4]x1
POL(rec(x1)) = [2]x1
POL(sent(x1)) = [4]x1
POL(up(x1)) = [3] + x1

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:

SENT(up(z0)) → c5(SENT(z0))
K tuples:

CHECK(no(z0)) → c14(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
NO(up(z0)) → c6(NO(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

### (19) 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.

SENT(up(z0)) → c5(SENT(z0))
We considered the (Usable) Rules:

check(no(z0)) → no(check(z0))
rec(bot) → up(sent(bot))
rec(no(z0)) → sent(rec(z0))
no(up(z0)) → up(no(z0))
check(no(z0)) → no(z0)
check(rec(z0)) → rec(check(z0))
rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
And the Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(CHECK(x1)) = [2]x1
POL(NO(x1)) = 0
POL(REC(x1)) = [4]x1
POL(SENT(x1)) = x1
POL(bot) = [1]
POL(c(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c10(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c13(x1, x2)) = x1 + x2
POL(c14(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(check(x1)) = [2]x1
POL(no(x1)) = [2]x1
POL(rec(x1)) = [4]x1
POL(sent(x1)) = [2]x1
POL(up(x1)) = [1] + x1

### (20) Obligation:

Complexity Dependency Tuples Problem
Rules:

rec(rec(z0)) → sent(rec(z0))
rec(sent(z0)) → sent(rec(z0))
rec(no(z0)) → sent(rec(z0))
rec(bot) → up(sent(bot))
sent(up(z0)) → up(sent(z0))
check(sent(z0)) → sent(check(z0))
check(rec(z0)) → rec(check(z0))
check(no(z0)) → no(check(z0))
check(no(z0)) → no(z0)
no(up(z0)) → up(no(z0))
Tuples:

REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
NO(up(z0)) → c6(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
CHECK(no(z0)) → c14(NO(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
S tuples:none
K tuples:

CHECK(no(z0)) → c14(NO(z0))
CHECK(up(z0)) → c10(CHECK(z0))
CHECK(no(z0)) → c13(NO(check(z0)), CHECK(z0))
REC(sent(z0)) → c1(SENT(rec(z0)), REC(z0))
CHECK(sent(z0)) → c11(SENT(check(z0)), CHECK(z0))
REC(rec(z0)) → c(REC(z0))
CHECK(rec(z0)) → c12(CHECK(z0))
NO(up(z0)) → c6(NO(z0))
REC(no(z0)) → c2(SENT(rec(z0)), REC(z0))
REC(up(z0)) → c4(REC(z0))
SENT(up(z0)) → c5(SENT(z0))
Defined Rule Symbols:

rec, sent, check, no

Defined Pair Symbols:

REC, SENT, NO, CHECK

Compound Symbols:

c1, c2, c4, c5, c6, c10, c11, c13, c14, c, c12

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

The set S is empty