(0) 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:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(0, z0) → c
-'(z0, 0) → c1
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
P(0) → c3
P(s(z0)) → c4
S tuples:
-'(0, z0) → c
-'(z0, 0) → c1
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
P(0) → c3
P(s(z0)) → c4
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-', P
Compound Symbols:
c, c1, c2, c3, c4
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
P(0) → c3
-'(z0, 0) → c1
-'(0, z0) → c
P(s(z0)) → c4
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
S tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
S tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(z0)) → z0
Tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
S tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
-'(
z0,
s(
z1)) →
c2(
-'(
z0,
p(
s(
z1)))) by
-'(x0, s(z0)) → c2(-'(x0, z0))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(z0)) → z0
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
S tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(11) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
p(s(z0)) → z0
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
S tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
-'
Compound Symbols:
c2
(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
-'(x0, s(z0)) → c2(-'(x0, z0))
We considered the (Usable) Rules:none
And the Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(-'(x1, x2)) = x2
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
S tuples:none
K tuples:
-'(x0, s(z0)) → c2(-'(x0, z0))
Defined Rule Symbols:none
Defined Pair Symbols:
-'
Compound Symbols:
c2
(15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(16) BOUNDS(1, 1)