Link Search Menu Expand Document

Propositions

We start with the basic building blocks of first-order logic: propositions. The collection of propositions is simply a collection of symbols. Common ones are based on the Roman or Greek alphabets, e.g. A,B,C,α,β,γ,A, B, C ,\ldots \\ \alpha, \beta, \gamma, \ldots But we could just as easily using emoji for our symbols

😉, 🤯, 🌭, . . .

Right now, they have no meaning attached.

Given a set of propositions, we next need rules to combine them into expressions.

Deduction

The operations we use to combine expressions will reflect oft-occuring structures in human reason. But, we also need to model the common steps in deduction. Here there are two separate roles: assumptions we are given and goals we want to establish (if we can).

In the written (or natural) language, you might see.

Example. If nn and mm are even integers, then so is nmn-m.

Recall here that the integers are Z={,2,1,0,1,2,}\mathbb{Z} = \lbrace \ldots, -2,-1,0,1,2, \ldots \rbrace

This could also be written as

Example. Let nn and mm be even integers. The integer nmn-m is also an even integer.

While the latter example is not written an explicit if-then the content is the same. Our set of assumptions is nn and mm are even integers while our goal is nmn-m is an even integer.

Let’s write down a proof of these statements. We can represent as a table.

AssumptionsGoal
n,mn, m evennmn-m is even

Mathematics is well-known for compact (and difficult) notation. In a proof, it is often valuable to unfold the definitions into the statement they represent. We perform this operation on the assumptions.

AssumptionsGoal
n=2cn = 2c for some integer ccnmn-m is even
m=2dm = 2d for some integer dd 

We also rewrite the goal in the same way.

AssumptionsGoal
n=2cn = 2c for some integer ccnm=2en-m = 2e for some integer ee
m=2dm = 2d for some integer dd 

To achieve our goal, we want to exhibit the desired ee using the assumptions at hand. To make progress, we import another “known fact” into the assumptions context: distribution of multiplication over subtraction.

AssumptionsGoal
n=2cn = 2c for some integer ccnm=2en-m = 2e for some integer ee
m=2dm = 2d for some integer dd 
u(vw)=uvuwu(v-w) = uv-uw for all integers u,v,wu,v,w 

Now we can an apply distribution with u=2,v=c,w=du = 2, v = c, w = d to gain another statement and finish.

AssumptionsGoal
n=2cn = 2c for some integer ccnm=2en-m = 2e for some integer ee
m=2dm = 2d for some integer dd 
u(vw)=uvuwu(v-w) = uv-uw for all integers u,v,wu,v,w 
nm=2c2d=2(cd)n - m = 2c - 2d = 2(c-d) 
set e=cde = c-d 
nm=2en -m = 2e 

While we modeled this proof as a table, it is notationally simpler to represent deduction from assumption AA to goal BB (with some intermediate steps) vertically.

proof tree example

And we write ABA \vdash B to indicate that we can prove the goal(s) BB given the assumption(s) AA.