(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
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) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))
Tuples:
+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(0) → c3
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(0, z0) → c6
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
S tuples:
+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(0) → c3
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(0, z0) → c6
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
K tuples:none
Defined Rule Symbols:
+, minus, *
Defined Pair Symbols:
+', MINUS, *'
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
MINUS(0) → c3
+'(0, z0) → c
*'(0, z0) → c6
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
S tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
K tuples:none
Defined Rule Symbols:
+, minus, *
Defined Pair Symbols:
+', MINUS, *'
Compound Symbols:
c1, c2, c4, c5, c7, c8
(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
We considered the (Usable) Rules:none
And the Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*(x1, x2)) = [3] + [4]x2
POL(*'(x1, x2)) = [5]x1
POL(+(x1, x2)) = [2] + [2]x1
POL(+'(x1, x2)) = [4]
POL(0) = [3]
POL(MINUS(x1)) = [3]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c8(x1, x2, x3)) = x1 + x2 + x3
POL(minus(x1)) = [3]
POL(p(x1)) = [4] + x1
POL(s(x1)) = [4] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
S tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
K tuples:
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
Defined Rule Symbols:
+, minus, *
Defined Pair Symbols:
+', MINUS, *'
Compound Symbols:
c1, c2, c4, c5, c7, c8
(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
We considered the (Usable) Rules:none
And the Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*(x1, x2)) = 0
POL(*'(x1, x2)) = x1 + x1·x2 + [2]x12
POL(+(x1, x2)) = 0
POL(+'(x1, x2)) = [2]
POL(0) = 0
POL(MINUS(x1)) = [2] + x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c8(x1, x2, x3)) = x1 + x2 + x3
POL(minus(x1)) = 0
POL(p(x1)) = [2] + x1
POL(s(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
S tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
K tuples:
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
Defined Rule Symbols:
+, minus, *
Defined Pair Symbols:
+', MINUS, *'
Compound Symbols:
c1, c2, c4, c5, c7, c8
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
We considered the (Usable) Rules:
+(s(z0), z1) → s(+(z0, z1))
*(p(z0), z1) → +(*(z0, z1), minus(z1))
minus(0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
minus(s(z0)) → p(minus(z0))
+(p(z0), z1) → p(+(z0, z1))
+(0, z0) → z0
And the Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*(x1, x2)) = x1·x2
POL(*'(x1, x2)) = x12·x2
POL(+(x1, x2)) = x1 + x2
POL(+'(x1, x2)) = x1
POL(0) = 0
POL(MINUS(x1)) = 0
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c8(x1, x2, x3)) = x1 + x2 + x3
POL(minus(x1)) = x1
POL(p(x1)) = [1] + x1
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
S tuples:none
K tuples:
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
Defined Rule Symbols:
+, minus, *
Defined Pair Symbols:
+', MINUS, *'
Compound Symbols:
c1, c2, c4, c5, c7, c8
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)