Solutions to the Quiz 4 (2016)
I601 Logic and Discrete Mathematics
A1.
Prove the validity of the formula ∀xP (x) ∨ ∀xQ(x) → ∀x∀y[P (x) ∨ Q(y)].
S o l u t i o n.
∀xP (x), P (z) ` P (z), Q(t)
∀xQ(x), Q(t) ` P (z), Q(t)
(∀L)
(∀L)
∀xP (x) ` P (z), Q(t)
∀xQ(x) ` P (z), Q(t)
(∨R)
(∨R)
∀xP (x) ` P (z)∨Q(t)
∀xQ(x) ` P (z)∨Q(t)
(∀R)
(∀R)
∀xP (x) ` ∀y[P (z) ∨ Q(y)]
∀xQ(x) ` ∀y[P (z) ∨ Q(y)]
(∀R)
(∀R)
∀xP (x) ` ∀x∀y[P (x) ∨ Q(y)]
∀xQ(x) ` ∀x∀y[P (x) ∨ Q(y)]
(∨L)
∀xP (x)∨∀xQ(x) ` ∀x∀y[P (x) ∨ Q(y)]
(→R)
` ∀xP (x) ∨ ∀xQ(x)→∀x∀y[P (x) ∨ Q(y)]
The formula is provable i.e. tautology.
B1.
Prove the validity of the formula ∀x∀y[P (x) ∨ Q(y)] → ∀xP (x) ∨ ∀xQ(x).
S o l u t i o n.
A, B(z), P (z) ` P (z), Q(t)
A, B(z), Q(t) ` P (z), Q(t)
A, B(z), P (z)∨Q(t) ` P (z), Q(t)
(∀L)
A, ∀y[P (z) ∨ Q(y)] ` P (z), Q(t)
(∀L)
∀x∀y[P (x) ∨ Q(y)] ` P (z), Q(t)
(∀R)
∀x∀y[P (x) ∨ Q(y)] ` P (z), ∀xQ(x)
(∀R)
∀x∀y[P (x) ∨ Q(y)] ` ∀xP (x), ∀xQ(x)
(∨R)
∀x∀y[P (x) ∨ Q(y)] ` ∀xP (x)∨∀xQ(x)
(→R)
` ∀x∀y[P (x) ∨ Q(y)]→∀xP (x) ∨ ∀xQ(x)
(∨L)
Here A and B(z) denote a proposition ∀x∀y[P (x) ∨ Q(y)] a predicate ∀y[P (z) ∨ Q(y)], respectively.
The formula is provable i.e. tautology.
A2.
Prove the identity
1 + 3 + 9 + 27 + · · · + 3n−1 =
3n − 1
.
2
S o l u t i o n.
Base Case: For n = 1, we have
31 − 1
2
= = 1.
2
2
Inductive Step: Assume, that 1 + 3 + 9 + 27 + · · · + 3k−1 =
k+1
k first multiples of three equals to 3 2 −1 :
1 + 3 + 9 + 27 + · · · + 3k−1 + 3k =
B2.
3k −1
2 ,
and prove that the sum of
3k − 1
+ 3k =
2
=
3k − 1 + 2 · 3k
=
2
=
3 · 3k − 1
=
2
=
3k+1 − 1
2
Prove that the sum of the first n squares (1 + 4 + 9 + · · · + n2 ) is n(n + 1)(2n + 1)/6.
S o l u t i o n.
Base Case: For n = 1, we have
1 · (1 + 1) · (2 · 1 + 1)
1·2·3
6
=
= =1
6
6
6
Inductive Step: Assume, that 1 + 4 + 9 + · · · + k 2 = k(k + 1)(2k + 1)/6, we have and prove
that the sum of k + 1 first perfect squares equals to (k + 1)(k + 2)(2k + 3) = 6:
1 + 4 + 9 + · · · + k 2 + (k + 1)2 =
k(k + 1)(2k + 1)
+ (k + 1)2 =
6
=
(k + 1)[k(2k + 1) + 6(k + 1)]
=
6
=
(k + 1)(2k 2 + 7k + 6)
=
6
=
(k + 1)(2k 2 + 3k + 4k + 6)
=
6
=
(k + 1)[k(2k + 3) + 2(2k + 3)]
=
6
=
(k + 1)(k + 2)(2k + 3)
6