(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
Tuples:
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
='(nil, nil) → c3
='(.(z0, z1), nil) → c4
='(nil, .(z0, z1)) → c5
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))
S tuples:
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
='(nil, nil) → c3
='(.(z0, z1), nil) → c4
='(nil, .(z0, z1)) → c5
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))
K tuples:none
Defined Rule Symbols:
del, f, =
Defined Pair Symbols:
DEL, F, ='
Compound Symbols:
c, c1, c2, c3, c4, c5, c6
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
='(nil, .(z0, z1)) → c5
='(nil, nil) → c3
='(.(z0, z1), nil) → c4
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
Tuples:
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
S tuples:
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
K tuples:none
Defined Rule Symbols:
del, f, =
Defined Pair Symbols:
DEL, F
Compound Symbols:
c, c1, c2
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
Tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
S tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
K tuples:none
Defined Rule Symbols:
del, f, =
Defined Pair Symbols:
F, DEL
Compound Symbols:
c1, c2, c
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
Tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
S tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
K tuples:none
Defined Rule Symbols:
=
Defined Pair Symbols:
F, DEL
Compound Symbols:
c1, c2, c
(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.
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
We considered the (Usable) Rules:
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
=(nil, .(z0, z1)) → false
=(.(z0, z1), nil) → false
=(nil, nil) → true
And the Tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(.(x1, x2)) = [4] + x2
POL(=(x1, x2)) = [5]
POL(DEL(x1)) = [2] + [4]x1
POL(F(x1, x2, x3, x4)) = [4]x1 + [4]x4
POL(and(x1, x2)) = [1]
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(false) = [5]
POL(nil) = [2]
POL(true) = [5]
POL(u) = [2]
POL(v) = 0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))
Tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
S tuples:none
K tuples:
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))
Defined Rule Symbols:
=
Defined Pair Symbols:
F, DEL
Compound Symbols:
c1, c2, c
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(O(1), O(1))