Outline
- Semantics of First-order Logic
- In-Class Exercise
- First-order Logic Proofs
Introduction
- On Monday, we started talking about First-Order Logic.
- When we were talking about search we switched from problem solving agents which worked on atomic domains,
to constraint satisfaction problems where the domain was factored.
- In similar fashion, we have talked about logic-based agents and we have said that one way to represent their knowledge
bases is with propositional logic.
- This works well for atomic domains, but now we are interested in logic-based agents that can work with factored domains, where
we have real world objects whose properties can take on more than binary values.
- First-order logic is our approach for doing this.
- In first-order logic, we have constants and variables which run over domains. We can build up terms using these, and the functions
of our first-order logic. We can substitute terms into the predicates of our logic to make atomic formulas, and we can build general first-order formulas out of these using `neg, ^^, vv, forall x, exists y`.
- Today, we start with a definition of the semantics of first-order logic.
Semantics 1st order formulas
- To start we need to have a set `M` called the universe that our variables range over.
- For each constant `c` in our language, we need to give a value `c^M` in `M`.
- Here the superscript means in the model, not exponentiation.
- For each function `f` in our language need to give an actual function
`f^M: M times ... times M -> M` determined on `M`
- For each predicate symbol `P`, `P^M:M times ... times M -> {T, F}`
- All of the above information together is called a structure/model. Usually, denoted by `M`.
- We can almost now talk about the meaning of a formula `F` in a given structure `M`...
- Consider:
`(exists x)G(x, y)`
in this formula `x` is called a bound variable as it is in the scope of an `exists` or `forall`, `y` is unbound.
- To give the meaning of a formula, we also need an unbound variable assignment. Here variable/object assignment `nu` is a map from unbound variables to elements of `M`. We can finally write `M |== F[v]` (`M` entails `F` with assignment `nu`)
What is the meaning of a formula `F`?
- If `F` is an atomic formula calculate the value in `M` of each of the terms in `F`. Plug these values into the predicate `P^M` and see if outputs `T` or `F`.
- If `F:= neg G`, then `F` is true in `M,nu` iff not `M |== G[nu]`.
- If `F:= G vv H`, then `M|== F[nu]` iff `M|== G[nu]` or `M|== H[nu]`
- If `F:= G ^^ H` then `M|== F[nu]` iff `M|== G[nu]` and `M|== H[nu]`
- Other connectives are similar.
- If `F := (exists x) G` then `M |== F[nu]` iff there is some way to map `x -> x^M in M` such that `nu` is extended by this additional mapping give `M |== G [x-> x^M, v]`
- If `F := (forall x) G` then `M |== F[nu]` iff for all ways to map `x -> x^M in M`, `M|== G[x->x^M , nu]`.
- Write `M |== F` if for every variable assignment, `nu, M |== F[v]`
Example
- Suppose we have the language: `0`,`1`, `+`, `cdot`, `=` consider the formula `(exists x) (1+1)cdot x = 1+1+1+1`.
- This intuitively asserts there is some number such that twice that number is four.
- All variables in this formula are bound.
- Let our universe `M` be the natural numbers, `NN`: 0,1,2,3 ...
- Recall we will use superscript to mean in the model of the natural numbers, not exponentiation.
- We interpret `0^M` to be 0, zero in the natural numbers, and `1^M` to be 1, one in the natural numbers.
- We interpret the functions `+^M` and `cdot^M` as the usual plus and times on the natural numbers.
- We interpret the predicate `=^M` as the usual equality of two natural numbers.
- From the last slide `M |== (exists x) (1+1)cdot x = 1+1+1+1` iff there is some variable `x -> x^M in M` such that
`M|== (1+1)cdot x = 1+1+1+1[x -> x^M]`.
- Suppose we map `x` to `2^M`. Then the statement holds if `(1^M +^M 1^M) cdot 2^M =^M 1^M +^M 1^M +^M 1^M +^M 1^M`. Now both sides, after computing the values in the model based on `1^M`, `2^M`, and `+^M`, are `4^M`. Thus, the statement does hold for our model.
- Consider the statement `(exists x) (1+1)cdot x = 1+1+1`. In our natural number model there is no assignment to `x->x^M` that could make this statement true. so `M` does not model `(exists x) (1+1)cdot x = 1+1+1`
- Consider the model `M'` with universe `{0,1,2,3,4}` and where we interpret `+` and `cdot` using the integers mod `5`. In this model both, `(exists x) (1+1)cdot x = 1+1+1+1` and `(exists x) (1+1)cdot x = 1+1+1` will hold. To see this for the second statement, notice that `2 cdot 4 mod5 equiv 3` This shows that a statement can be true in more than one model and it also shows just because a statement is false in one model doesn't mean it is false in all models.
In-Class Exercise
- Let our language consist of the predicate `Parent(x,y)`.
- Consider the formula `(forall x) neg Parent(x,x) ^^ (exists y) Parent(z,y)`.
- Give a model and variable assignment under which this statement is true.
- Post your solutions to the Oct 25 In-Class Exercise Thread.
KBs, First-Order, Proofs
- Write `KB |== F` to mean for all structures `M`, such that `M |== KB`
we also have `M |== F`
- Proofs in 1st Order Logic are
very much like proofs in propositional logic except now we have some additional axioms.
- For example, some possible additional axioms might be:
`(forall x) (neg P) => neg((exists x) P)`
(for every pig, it can't fly = not there is a pig that can fly)
`neg(forall x P) =>(exists x) neg P`
(Not for every horse it is blue = there is a horse that is not blue)
`A(t) => (exists x) A(x)`
`A(x)=> (forall y) A(y)`
etc.
