Answer
$a)∀x∀yP(x,y):
P(1,1)∧P(1,2)∧P(1,3)∧P(2,1)∧P(2,2)∧P(2,3)∧P(3,1)∧P(3,2)∧P(3,3)$
$b)∃x∃yP(x,y):
P(1,1)∨P(1,2)∨P(1,3)∨P(2,1)∨P(2,2)∨P(2,3)∨P(3,1)∨P(3,2)∨P(3,3)$
$c)∃x∀yP(x,y):
(P(1,1)∧P(1,2)∧P(1,3))∨(P(2,1)∧P(2,2)∧P(2,3))∨(P(3,1)∧P(3,2)∧P(3,3))$
$d)∀y∃xP(x,y):
(P(1,1)∨P(2,1)∨P(3,1))∧(P(1,2)∨P(2,2)∨P(3,2))∧(P(1,3)∨P(2,3)∨P(3,3))$
Work Step by Step
Initial data:
The domain of $P(x,y)P(x, y)$ consists of pairs where $(x,y)(x, y)$ :
$x∈{1,2,3} \in \{1, 2, 3\}$
$y∈{1,2,3} \in \{1, 2, 3\}$
The propositions will be expressed using disjunctions $(∨)$ and conjunctions $(∧)$.
a) $∀x∀yP(x,y):$
This proposition asserts that $P(x, y)$ is true for all pairs $(x, y)$ in the domain.
Step-by-step instructions:
1. We expand the universal quantifier $∀𝑥$: this is equivalent to considering all possible values of $x: 𝑥=1,2,3x=1,2,3.$
2. For each value of $x$, we expand $∀y$, considering all possible values of $y: 𝑦=1,2,3$
3. For $∀𝑥∀y𝑃(𝑥,𝑦)$ to be true, we must have the conjunction $(∧)$ of $𝑃(𝑥,𝑦)$ for all pairs.
Answer:
$$P(1,1)∧P(1,2)∧P(1,3)∧P(2,1)∧P(2,2)∧P(2,3)∧P(3,1)∧P(3,2)∧P(3,3)$$
b) $∃𝑥∃𝑦𝑃(𝑥,𝑦)$:
This proposition states that there is at least one pair $(x,y)$ for which $𝑃(𝑥,𝑦)$ is true.
Step by step instructions:
1.We expand the existential quantifier $∃𝑥$: this means that we consider that $x$ can be $1,2,or, 3$, but we only need valid $x$ .
2. For each $x$, we expand $∃𝑦$: this means that we consider that $y $ can be $1,2,or,3$, but we only need one valid $y$ .
3. For $∃𝑥∃𝑦𝑃(𝑥,𝑦)$ to be true, it is enough to have the disjunction $(
∨)$ of $𝑃(𝑥,𝑦)$ for all pairs.
Answer:
$$P(1,1)∨P(1,2)∨P(1,3)∨P(2,1)∨P(2,2)∨P(2,3)∨P(3,1)∨P(3,2)∨P(3,3)$$
c)$∃x∀yP(x,y)$:
This proposition states that there is a value of $𝑥$ such that $𝑃(𝑥,𝑦)$ is true for all values of $𝑦$
Step by step instructions:
1. We expand the existential quantifier $∃x$: this means that there is at least one value of $𝑥$ between $1,2,3$ that satisfies the condition.
2. For that specific value of $𝑥$, we expand $∀y$: this means that $𝑃(𝑥,𝑦)$
must be true for all values of $y$.
3. For $∃𝑥∀𝑦𝑃(𝑥,𝑦)$ to be true, we take the disjunction $(∨)$ over the values of $𝑥$, and for each $𝑥$, the conjunction $(∧)$ over the values of $𝑦$.
Answer:
$$(P(1,1)∧P(1,2)∧P(1,3))∨(P(2,1)∧P(2,2)∧P(2,3))∨(P(3,1)∧P(3,2)∧P(3,3))$$
d)$∀y∃xP(x,y)$:
This proposition states that for each value of $y$ ,there is at least one
$𝑥$ such that$𝑃(𝑥,𝑦)$ is true.
Step by step instructions:
1. We expand the universal quantifier $∀𝑦$: this means that we consider all the values of $y: y=1,2,3.$
2. For each value of $y$ , we expand the existential quantifier $∃x$: this means that for each $y$, there must be at least a value of $x$ between $1,2,3$ for which $P(x,y)$ is true.
3. For $∀y∃xP(x,y)$ to be true, we take the conjunction $(∧)$ over the values of $y$, and for each $y$, the disjunction $(∨)$ over the values of $x$.
Answer:
$$(P(1,1)∨P(2,1)∨P(3,1))∧(P(1,2)∨P(2,2)∨P(3,2))∧(P(1,3)∨P(2,3)∨P(3,3))$$
