Logic for Computer Scientists/Predicate Logic/Semantics

Definition 4 (Semantics of predicate logic - Interpretation)
An interpretation is a pair $$\mathcal{I} = (U_\mathcal{I}, A_\mathcal{I})$$, where
 * $$U_\mathcal{I}$$ is an arbitrary nonempty set, called domain, or universe.
 * $$A_\mathcal{I}$$ is a mapping which associates to
 * a $$k$$-ary predicate symbol, a $$k$$-ary predicate over $$U_\mathcal{I}$$,
 * a $$k$$-ary function symbol, a $$k$$-ary function over $$U_\mathcal{I}$$, and
 * a variable an element from the domain.

Let $$F$$ be a formula and $$\mathcal{I} = (U_\mathcal{I}, A_\mathcal{I})$$ be an interpretation. We call $$\mathcal{I}$$ an interpretation for $$F$$, if $$\mathcal{A}_\mathcal{I}$$ is defined for every predicate and function symbol, and for every variable, that occurs free in $$F$$.

Example: Let $$F= \forall x p(x,f(x)) \land q(g(a,z))$$ and assume the varieties of the symbols as written down. In the following we give two interpretations for $$F$$: For a given interpretation $$\mathcal{I}= (U,A)$$ we write in the following $$p^\mathcal{I}$$ instead of $$A(p)$$; the same abbreviation will be used for the assignments for function symbols and variables.
 * $$\mathcal{I}_1 = (N_0,A_1)$$, such that
 * $$A_1(p) = \{ (m,n) \mid m,n \in N_0 \text{ and } m < n\}$$
 * $$A_1(q) = \{ n \in N_0 \mid n \text{ is prime} \}$$
 * $$A_1(f)(n) = n+1 \; \forall n \in N_0$$
 * $$A_1(g)(m,n) = m+n \; \forall n,m \in N_0$$
 * $$A_1(a) = 2$$
 * $$A_1(z) = 3$$
 * Under this interpretation the formula $$F$$ can be read as " Every natural number is smaller than its successor and the sum of $$2$$ and $$3$$ is a prime number."
 * $$\mathcal{I}_2 = (U_2,A_2)$$, such that
 * $$U_2= \{a, f(a), g(a,a), f(g(a,a)), g(f(a),f(a)), \cdots \}$$
 * $$A_2(f)(t) = f(t)$$ for $$t \in U_2$$
 * $$A_2(g)(t_1,t_2) = g(t_1,t_2)$$, if $$t_1,t_2 \in U_2$$
 * $$A_2(a) = a$$
 * $$A_2(z)= f(f(a))$$
 * $$A_2(p)=\{p(a,a), p(f(a),f(a)),p(f(f(a)),f(f(a)))\}$$
 * $$A_2(q)=\{g(t_1,t_2) \mid t_1,t_2 \in U_2\}$$

Definition 5 (Semantics of predicate logic - Evaluation of Formulae)
Let $$F$$ be a formula and $$\mathcal{I}$$ an interpretation for $$F$$. For terms $$t$$ which can be composed with symbols from $$F$$ the value $$\mathcal{I}(t)$$ is given by The value $$\mathcal{I}(F)$$ of a formula $$F$$ is given by \begin{cases} \;\;\,true & \text{ if } (\mathcal{I}(t_1), \cdots, \mathcal{I}(t_k))\in p^\mathcal{I} \\ \;\;\,false & \; otherwise \end{cases}$$
 * $$\mathcal{I}(x)= x^\mathcal{I}$$
 * $$\mathcal{I}(f(t_1, \cdots, t_k)) = f^\mathcal{I}(\mathcal{I}(t_1), \cdots , \mathcal{I}(t_k))$$, if $$t_1, \cdots, t_k$$ are terms and $$f$$ a $$k$$-ary function symbol. (This    holds for the case $$k=0$$ as well.)
 * $$ \mathcal{I}(p(t_1, \cdots, t_k)) =

\begin{cases} \;\;\,true & \text { if } \mathcal{I}(F) = true { and } \mathcal{I}(G) = true \\ \;\;\,false& otherwise \end{cases} $$ \begin{cases} \;\;\,true & \text { if } \mathcal{I}(F) = true { or } \mathcal{I}(G) = true \\ \;\;\, false& otherwise \end{cases} $$
 * $$ \mathcal{I}(F\land G)) =
 * $$ \mathcal{I}((F\lor G)) =

\;\;\,true & \text { if } \mathcal{I}(F) = false \\ \;\;\, false& otherwise \end{cases} $$ \;\;\,true & \text{ if for every } d\in U\; : \; \mathcal{I}_{[x/d]}(G) = true \\ \;\;\, false& otherwise \end{cases} $$ \;\;\,true & \text{ if there is a }    d\in U\; : \; \mathcal{I}_{[x/d]}(G) = true \\ \;\;\, false& otherwise \end{cases} $$
 * $$ \mathcal{I}(\lnot F) = \begin{cases}
 * $$\mathcal{I}(\forall G) = \begin{cases}
 * $$\mathcal{I}(\exists G) = \begin{cases}

where, $$ f_{[x/d]}(y) = \begin{cases} \;\;\, f(y)& \text{ if }  x \neq y \\ \;\;\, d  &  otherwise \end{cases} $$

The notions of satisfiable, valid, and $$\models $$ are defined according to the propositional case (Semantic (Propositional logic)).

Note that, predicate calculus is an extension of propositional calculus: Assume only $$0$$-ary predicate symbols and a formula which contains no variable, i.e. there can be no terms and no quantifier in a well-formed formula.

On the other hand, predicate calculus can be extended: If one allows for quantifications over predicate and function symbols, we arrive at a second order predicate calculus. E.g. $$ \forall p \exists f \; p(f(x))$$ Another example for a second order formula of is the induction principle from Induction.

Problem 1 (Predicate)
The interpretation $$\mathcal{I} = \text{ is given } (U_\mathcal{I}, A_\mathcal{I}) $$ as follows: $$ U_\mathcal{I} =  \N $$ $$ p^\mathcal{I} = { (m,n) \mid m < n } $$ $$ f^\mathcal{I}(m,n) =  m+n $$ $$ x^\mathcal{I} = 5 ;  y^\mathcal{I} = 7 $$ Determine the value of following terms and formulae:
 * 1) $$\mathcal{I}(f(f(x,x),y))$$
 * 2) $$\mathcal{I}(\forall x \forall y (p(x,y) \lor p(y,x))$$
 * 3) $$\mathcal{I}(p(x,x) \to p(y,x))$$
 * 4) $$\mathcal{I}(\exists x p(y,x))$$

$$\Box$$

Problem 2 (Predicate)
The interpretation $$\mathcal{I} = \text{ is given } (U_\mathcal{I},A_\mathcal{I})$$ as follows: $$ U_\mathcal{I} =  \R $$ $$ P^\mathcal{I} =  { z \mid z \geq 0 } $$ $$ f^\mathcal{I}(z) =  z^2  $$ $$ x^\mathcal{I} = \sqrt{2} $$ $$ E^\mathcal{I} =  { (x,y) \mid x=y } $$ $$ g^\mathcal{I}(x,y) =  x+y  $$ $$ y^\mathcal{I} =  -1 $$ Determine the value of following terms and formulae:

$$ 1. \mathcal{I}(g(f(x),f(y)))$$ $$ 2. \mathcal{I}(\forall x\,P(f(x)))$$ $$3. \mathcal{I}(\exists z \forall x \forall y\,E(g(x,y),z)) $$ $$ 4. \mathcal{I}(\forall y (E(f(x),y) \to P(g(x,y)))) $$

$$\Box$$

Problem 3 (Predicate)
The following formula is given: $$ F= \forall x \forall y \forall z\,R(h(h(x,y),z),h(x,h(y,z))) \ \land\ \exists x \exists y\,\lnot R(h(x,y),h(y,x)) $$ Indicate a structure $$\mathcal{A} $$, which is a model for $$F$$ and a structure $$\mathcal{B}$$ which is no model for $$F$$! $$\Box$$