(0) 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:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
lt(z0, 0) → false
lt(0, s(z0)) → true
lt(s(z0), s(z1)) → lt(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
div(z0, 0) → 0
div(0, z0) → 0
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:
-'(z0, 0) → c
-'(0, s(z0)) → c1
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(z0, 0) → c3
LT(0, s(z0)) → c4
LT(s(z0), s(z1)) → c5(LT(z0, z1))
IF(true, z0, z1) → c6
IF(false, z0, z1) → c7
DIV(z0, 0) → c8
DIV(0, z0) → c9
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
-'(z0, 0) → c
-'(0, s(z0)) → c1
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(z0, 0) → c3
LT(0, s(z0)) → c4
LT(s(z0), s(z1)) → c5(LT(z0, z1))
IF(true, z0, z1) → c6
IF(false, z0, z1) → c7
DIV(z0, 0) → c8
DIV(0, z0) → c9
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:
-, lt, if, div
Defined Pair Symbols:
-', LT, IF, DIV
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 8 trailing nodes:
DIV(z0, 0) → c8
-'(z0, 0) → c
DIV(0, z0) → c9
IF(false, z0, z1) → c7
LT(0, s(z0)) → c4
IF(true, z0, z1) → c6
-'(0, s(z0)) → c1
LT(z0, 0) → c3
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
lt(z0, 0) → false
lt(0, s(z0)) → true
lt(s(z0), s(z1)) → lt(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
div(z0, 0) → 0
div(0, z0) → 0
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:
-, lt, if, div
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
lt(z0, 0) → false
lt(0, s(z0)) → true
lt(s(z0), s(z1)) → lt(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
div(z0, 0) → 0
div(0, z0) → 0
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:
-, lt, if, div
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
lt(z0, 0) → false
lt(0, s(z0)) → true
lt(s(z0), s(z1)) → lt(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
div(z0, 0) → 0
div(0, z0) → 0
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:
-
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(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.
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
We considered the (Usable) Rules:
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
-(z0, 0) → z0
And the Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(-(x1, x2)) = x1
POL(-'(x1, x2)) = 0
POL(0) = 0
POL(DIV(x1, x2)) = x1
POL(LT(x1, x2)) = 0
POL(c10(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
K tuples:
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
Defined Rule Symbols:
-
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(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.
-'(s(z0), s(z1)) → c2(-'(z0, z1))
We considered the (Usable) Rules:
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
-(z0, 0) → z0
And the Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(-(x1, x2)) = x1
POL(-'(x1, x2)) = x2
POL(0) = 0
POL(DIV(x1, x2)) = x1·x2
POL(LT(x1, x2)) = 0
POL(c10(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:
LT(s(z0), s(z1)) → c5(LT(z0, z1))
K tuples:
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
-'(s(z0), s(z1)) → c2(-'(z0, z1))
Defined Rule Symbols:
-
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
LT(s(z0), s(z1)) → c5(LT(z0, z1))
We considered the (Usable) Rules:
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
-(z0, 0) → z0
And the Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(-(x1, x2)) = x1
POL(-'(x1, x2)) = 0
POL(0) = [1]
POL(DIV(x1, x2)) = [2]x1 + [2]x1·x2
POL(LT(x1, x2)) = x2
POL(c10(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(z0, 0) → z0
-(0, s(z0)) → 0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:none
K tuples:
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
-'(s(z0), s(z1)) → c2(-'(z0, z1))
LT(s(z0), s(z1)) → c5(LT(z0, z1))
Defined Rule Symbols:
-
Defined Pair Symbols:
-', LT, DIV
Compound Symbols:
c2, c5, c10
(15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(16) BOUNDS(1, 1)