Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.18)

The rewrite relation of the following TRS is considered.

sum(0) → 0 (1)
sum(s(x)) → +(sum(x),s(x)) (2)
+(x,0) → x (3)
+(x,s(y)) → s(+(x,y)) (4)
The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n2).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:
sum#(0) → c (5)
originates from
sum(0) → 0 (1)
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
originates from
sum(s(z0)) → +(sum(z0),s(z0)) (6)
+#(z0,0) → c2 (9)
originates from
+(z0,0) → z0 (8)
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)
originates from
+(z0,s(z1)) → s(+(z0,z1)) (10)
Moreover, we add the following terms to the innermost strategy.
sum#(0)
sum#(s(z0))
+#(z0,0)
+#(z0,s(z1))

1.1 Rule Shifting

The rules
sum#(0) → c (5)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3(x1)] = 1 · x1 + 0
[sum(x1)] = 2 + 2 · x1
[+(x1, x2)] = 3 · x2 + 0
[sum#(x1)] = 3
[+#(x1, x2)] = 0
[0] = 1
[s(x1)] = 0
which has the intended complexity. Here, only the following usable rules have been considered:
sum#(0) → c (5)
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
+#(z0,0) → c2 (9)
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)

1.1.1 Rule Shifting

The rules
+#(z0,0) → c2 (9)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3(x1)] = 1 · x1 + 0
[sum(x1)] = 3
[+(x1, x2)] = 3 + 3 · x2
[sum#(x1)] = 1 · x1 + 0
[+#(x1, x2)] = 1
[0] = 3
[s(x1)] = 1 + 1 · x1
which has the intended complexity. Here, only the following usable rules have been considered:
sum#(0) → c (5)
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
+#(z0,0) → c2 (9)
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)

1.1.1.1 Rule Shifting

The rules
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3(x1)] = 1 · x1 + 0
[sum(x1)] = 1 + 1 · x1
[+(x1, x2)] = 1 + 1 · x2
[sum#(x1)] = 1 · x1 + 0
[+#(x1, x2)] = 0
[0] = 1
[s(x1)] = 1 + 1 · x1
which has the intended complexity. Here, only the following usable rules have been considered:
sum#(0) → c (5)
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
+#(z0,0) → c2 (9)
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)

1.1.1.1.1 Rule Shifting

The rules
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)
are strictly oriented by the following non-linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3(x1)] = 1 · x1 + 0
[sum(x1)] = 2 + 2 · x1 + 2 · x1 · x1
[+(x1, x2)] = 1 + 1 · x2 + 2 · x2 · x2
[sum#(x1)] = 1 · x1 · x1 + 0
[+#(x1, x2)] = 1 · x2 + 0
[0] = 0
[s(x1)] = 1 + 1 · x1
which has the intended complexity. Here, only the following usable rules have been considered:
sum#(0) → c (5)
sum#(s(z0)) → c1(+#(sum(z0),s(z0)),sum#(z0)) (7)
+#(z0,0) → c2 (9)
+#(z0,s(z1)) → c3(+#(z0,z1)) (11)

1.1.1.1.1.1 R is empty

There are no rules in the TRS R. Hence, R/S has complexity O(1).