Mathematical Proof and the Principles of Mathematics/Logic/Logical equivalence

There is a final logical connective we'll be covering.


Again, we 'define' this in terms of other connectives and verify that it matches the definition given earlier. In this case we take

(  implies  ) and (  implies  )

to be the definition of

  iff  .

This definition is where the 'if and only if' expression comes from, since


can be phrased   if   and


can be phrased as   only if  .

So see that this matches the previous definition, notice that   implies   is False when   is True and   is False, meanwhile   implies   is False when   is True and   is False, so the expression (  implies  ) and (  implies  ) can only be true when   and   are both True or both False.

By combining this definition with the other rules of inference we get the following:

From   implies  ,   implies   deduce   iff  .
If by making the assumption not   one can derive  , and if by making the assumption not   one can derive   then deduce   iff  .
From  ,   iff   deduce  .
From  ,   iff   deduce  .

So you can use equivalence like implication except it goes in both directions, and to proof an equivalence is the same as proving an implication except you do it in both directions.

Example Proof #1Edit

Prop 1:   or   iff   or  

In this case we already have   or   implies   or   as Prop. 1 on the previous page. By switching the letters   with   in the proposition we get   or   implies   or  . So the proof is just a matter of putting these two previous results together.

Line Statement Justification
1   or   implies   or   Prop. 1 from the previous page.
2   or   implies   or   Also Prop. 1 from the previous page.
3   or   iff   or   From 1 and 2

Example Proof #2Edit

For a more complicated example, we'll need to use subproofs.

Prop 2:   or   iff (  implies  ) implies  

We're proving two implications, one in each direction, so the structure of the proof looks like:

Line Statement Justification
1   or   Hypothesis
n (  implies  ) implies   ?
n+1 (  implies  ) implies   Hypothesis
n+p   or   ?
n+p+1   or   iff (  implies  ) implies   From 1 and n, n+1 and n+p

At this point we've broken the proof into parts which can be tackled separately. In the first case we need to prove an implication, but it seems easier to do the usual way. In the second case we need to prove disjunction and there are two ways of doing this. It looks like the easier way is to assume not   and prove  . Now that we have several ways of proving things it may be that some ways are simpler than others, but without enough experience to get a feel for it you may have to use trial and error to find the simplest way. Keep in mind though, that when you see a proof in a book or article, the author may have gone through quite a few failed attempts before finding a proof that worked, but you don't get to see the failures.

Filling in more of the structure we have:

Line Statement Justification
1   or   Hypothesis
2   implies   Hypothesis
n−1   ?
n (  implies  ) implies   From 2 and n−1
n+1 (  implies  ) implies   Hypothesis
n+2 not   Hypothesis
n+p−1   ?
n+p   or   From n+2 and n+p−1
n+p+1   or   iff (  implies  ) implies   From 1 and n, n+1 and n+p

(To save space, we're constructing both halves of the proof at once; normally you'd do one at time though.)

In the first half we need to prove   and it appears that the best way to do this is using the method of contradiction. For the second half we already have enough results from previous pages to fill in the rest. As usual, we encourage you to try to fill in the rest yourself before looking at the final result. By different methods you may even find a simpler proof than the one given.

Line Statement Justification
1   or   Hypothesis
2   implies   Hypothesis
3 not   Hypothesis
4   From 1 and 3
5 not   From 2 and 3
6   From 4 and 5
7   From 3 and 6
8 (  implies  ) implies   From 2 and 7
9 (  implies  ) implies   Hypothesis
10 not   Hypothesis
11   implies   Prop. 8 on the page on indirect proofs.
12   From 9 and 11
13   or   From 10 and 12
14   or   iff (  implies  ) implies   From 1 and 8, 9 and 13

In prose form it's a good idea to label the statements and clearly mark which implication is being being proved when so that the reader can easily follow the proof.

Prop 2:   or   iff (  implies  ) implies  
Proof: We first prove that   or   implies ((  implies  ) implies  ). Assume   or  . Now assume   implies  . We need to proof  , so suppose not  . Then   by the first assumption and not   by the second assumption. This is a contradiction so we have  . Therefore (  implies  ) implies  . From the original assumption   or   it follows that (  implies  ) implies  , so we can conclude that   or   implies ((  implies  ) implies  ).
Now we prove that ((  implies  ) implies  ) implies   or  . Assume (  implies  ) implies  . Now assume not  . Then   implies   and therefore  , so   or  .

Left for the readerEdit

We've now covered nearly all of the commonly used rules of inference, so there is no shortage of statements that can be proved now. Some of the following are relatively trivial extensions of previous results, and some will require one or more subproofs, but it's up to you to figure out which is which.

Prop. 3: (  or  ) or   iff   or (  or  )
Prop. 4:   and   iff   and  
Prop. 5: (  and  ) and   iff   and (  and  )
Prop. 6: (  and  ) or   iff (  or  ) and (  or  )
Prop. 7: (  or  ) and   iff (  and  ) or (  and  ).
Prop. 8: (  iff  ) implies (  iff  )
Prop. 9: (  iff  ) and (  iff  ) implies (  iff  )
Prop. 10:   implies   iff not   or  
Prop. 11: not (  implies  ) iff   and not  
Prop. 11: not (  or  ) iff not   and not  
Prop. 12: not (  and  ) iff not   or not  
Prop. 13: (  iff  ) implies (  implies   iff   implies  )
Prop. 14: (  iff  ) implies (  implies   iff   implies  )
Prop. 15:   iff not not  

Groups of equivalent statementsEdit

In some cases a theorem may state that a group of several statements are equivalent to each other. For example the statement of the theorem might be in the form:

Theorem: The following are equivalent:
1. Statement 1
2. Statement 2
3. Statement 3
4. Statement 4

This says Statement 1 iff Statement 2, Statement 1 iff Statement 3, ... Statement 3 iff Statement 4, with all 6 possible variations. The usual way to proof this type of theorem is to prove implications in a cycle. In this case you would prove,

P1: Statement 1 implies Statement 2
P2: Statement 2 implies Statement 3
P3: Statement 3 implies Statement 4
P4: Statement 4 implies Statement 1

This is very efficient since by proving just four implications you've in effect proven all 12 possible implications between two of the four statements. The reason this works is by repeated application of the following

Prop. 13: (  implies  ) and (  implies  ) implies (  implies  )
Proof: (This is an easy application of previous results and is left as an exercise.)

As a further exercise, prove that

P1 and P2 and P3 and P4 implies (Statement 1 iff Statement 3)


When two statements are logically equivalent they have the same truth value. So it seems reasonable to claim that if   is equivalent to   and   is some expression involving  , then   is equivalent to  , where   is obtained from   by replacing the statement   by  .

As an example of how this might be applied, we have from Prop. 15 above that   is equivalent to not not  . It would be nice to conclude then that

not not   implies   or (  iff  )

is equivalent to

  implies   or (  iff  )

without going through a separate proof.

This is a valid conclusion, but the tools to justify it belong to the realm of mathematical logic and are outside the scope of this book. We can give proofs on a case by case basis though, and some examples serve to demonstrate how these proofs can be constructed. If the expression   involves only implication and the logical constant  , then we can apply repeatedly apply propositions 13 and 14 above.

For example, if   is

  implies (  implies  )

then we can prove

  implies (  implies  )

is equivalent to

  implies (not not   implies  )

as follows:

not not   iff   (by Prop. 15)
not not   implies   iff   implies   (by Prop. 13)
  implies (not not   implies  ) iff   implies (  implies  ) (by Prop. 14)

For an expression involving other logical connectives we can use the definitions we gave in terms of implication to turn the expression into one of the previous type.