M is a model with domain M(∀) = {1, 2, 3} and
• M(a) = 1, M(b) = 3, M(c) = 2
• M(P) = {1, 3}, M(R) = {(3, 1),(2, 2),(2, 1),(3, 3)} h is a variable assignment with h(x) = 2, h(y) = 3, and h(z) = 2.
For each of the following statements, determine whether it is true or false. Explain your answers by following the definition of satisfaction step by step.
I'm having trouble with the following two statements:
1) M |= R(c, y)[h[y/1]]
2) M |= c = x ∧ R(y, x)[h]
Appreciating all input.
It looks little less cluttered (and it's standard usage) to use $M$ as a superscript on symbols, indicating "the interpretation of (that symbol) in the model $M$". Thus, $c^M = M(c) = 2$, and $R^M = M(R)$, etc.
1) True
I take it that $h[y/1]$ means "the map which agrees with $h$ everywhere except at '$y$' where its value is $1$".
We need to check whether a particular ordered pair is in $R^M$, namely, whether or not $$ (c^M, h[y/1](y)) \in R^M $$ By definition, $h[y/1](y) = 1$, and $c^M = 2$, so 1) is true provided $$ (2,1) \in R^M $$ Checking your definition of $R^M = M(R)$, this is the case.
2) False
$M \models (c = x \wedge R(y,x))[h]$ is true iff both of the following are true: $$M \models (c = x)[h] \tag{2a} $$ and $$M \models R(y,x)[h] \tag{2b} \text{.} $$ (2a) is true iff $c^M = h(x)$; as $c^M = 2$ and $h(x) = 2$, they are equal.
(2b) is true iff $(h(y), h(x)) \in R^M$. But $(h(y), h(x)) = (3,2) \notin R^M$, so (2b) is false, thus 2) is false.