12/25/06
You now have to tools to appraise propositional arguments.
Let us run through how these might be used with two examples.
Consider the argument
If no human action is free, then no one is responsible for what they do.
If no one is responsible for what they do, no one should be punished.
Therefore
If no human action is free, no one should be punished.
First it should be symbolized
F = no human action is free
R = no one is responsible for what they do
P = no one should be punishedF⊃R, R⊃P ∴ F⊃P
Then some thought should be given as to whether there is a counter-example and thus whether the argument is invalid. In a counter example the conclusion has to be false
F ⊃ R R ⊃ P F ⊃ P F
but F⊃P is false only if F if true and P is false.
F ⊃ R R ⊃ P F ⊃ P T F F
and these assignments also have to apply to the premises
F ⊃ R R ⊃ P F ⊃ P T F T F F
but now we seem to be stuck for if R is true then the second premise is false but if R is false the first premise is false so it seems that we cannot find an assignment which renders all premises true and the conclusion false at one and the same time. It seems that this argument has no counter-example.
So now we try to give a derivation of it. And indeed we are able to find
1 F⊃R 2 R⊃P 3 F⊃P 1,2 HS
and this proves that the argument is valid.
Consider the argument
If the girl has spots in her mouth, then she has measles.
If the girl has spots on her back, then she has a heat rash.
The girl has spots on her back.
Therefore
The girl has a heat rash and does not have measles.
First it should be symbolized
S = the girl has spots in her mouth
M = the girl has measles
B = the girl has spots on her back
R = the girl has a heat rashS⊃M, B⊃R, B ∴ R.∼M
Then some thought should be given as to whether there is a counter-example and thus whether the argument is invalid. In a counter example all the premises have to be true and that tells us that B has to be assigned true
S ⊃ M B ⊃ R B R . ∼ M T
and this assignment applies to all other B's
S ⊃ M B ⊃ R B R . ~ M T T
the second premise B⊃R has to be true and given that B is true this means that R must be true (throughout the argument)
S ⊃ M B ⊃ R B R . ∼ M T T T T T
the conclusion R.M has to be false and given that R is true this must mean that ∼M is false and M true throughout the argument.
S ⊃ M B ⊃ R B R . ∼ M T T T T T T F F T
all that remains is S and the first premise S⊃M-- M is true and we wish to have the whole premise true that means that S has can be either true or false but, just to pick one, we will assign it true
S ⊃ M B ⊃ R B R . ∼ M T T T T T T T T F F T
This argument has a counter-example; the assignment B true, S true, M false and R true proves the argument to be invalid.
Logicians really need to keep practicing with derivations (just like musicians need to work on their scales).
[To make your practice more effective, the Derive It and Next Line functions will give you the next lines, but conceal where they have come from. You can think that through for yourself.]
You can, in this exercise, try any derivation you wish. Paste into the lower area whatever text you plan to use, then select a derivation and start a proof.
There are a few points to keep in mind when doing this
a) these are the logical symbols we work with
∼ . ∨ ⊃ ≡ ∴
and it is important to get them right. So, for example, A ∨ B is a perfectly good formula (with the real html or-symbol for the 'or' in the middle); but you might instead try to use upper or lower case letter 'vee' to make a formula ie A v B or A V B ; now, these may look the same to you, but the computer itself knows that you have used the wrong symbol and it will reject the formula (actually, to add a level of detail, the program is smart enough to correct you if it can, but upper case vee V is used for propositions and lower case vee v is used for variables in predicate logic (which you will meet shortly) so, more than likely, the program will be unable to correct you if you start using vees in the wrong place). If in doubt, copy and paste, or drag and drop from the above sample of logical symbols.
b) you need to select carefully and properly. With a formula like (A ∨ B) if you select (A ∨ B) all will be well, but if you select (A ∨ B) the program will reject it (you have missed a right hand bracket). These are easy slips to make.
c) when you are pasting, or dragging and dropping, from elsewhere, sometimes that action picks up some extra invisible symbols usually on the end. (What is happening here is usually that other programs tag or identify what you cut or copy or drag. For example, if you drag from a web page in a web browser, some browsers add a invisible tag to your piece that says 'this is a web page fragment'. Now, these tags may be invisible to you, but the computer can see them. And these invisible symbols can cause the program to reject a formula. Of course, the program knows all about existing tags and will ignore them. But this is a moving target, as new programs come out, new types of tag might appear.) There are two solutions to this relatively rare occurrence. First, and easiest, put your insertion point after the new text (ie click after the new text); then press the delete or backspace key; if that action removes the last visible character, nothing was there, and you should Undo your edit to restore where you where; if that action seems to do nothing at all, there was an invisible character there which you have just deleted, which is what you want -- hitting delete a second time will then remove the last visible character. Second, for geeks, look at the underlying html text in an html editor, and remove the noise.
Proofs
Here are some valid derivations (or use your own from elsewhere)
M⊃N, N⊃O,M ∴O
M⊃(N⊃O), M⊃N,M ∴O.M
M⊃((~N)⊃O), M⊃~N,M ∴O.M,
M≡N ∴ N⊃M
M≡N ∴ M⊃N
M≡N, N ∴ O∨M
M≡N, N≡O, M ∴ O
∼∼R∴R
R, ∼∼∼∼S,T∴S
R.S ∴ S
∼∼((R.S).(R.T)) ∴ R.S
R,(R.∼∼S).(R.T),T ∴ S
R,S ∴ R.S
R,S ∴ S.R
R.S,R,∼∼T ∴ (R.S).(R.T)
R,S,T ∴ (R.S).(R.T)
R ∴ R.R
R.(S.T) ∴ (R.S).T
(R.S).(R.T) ∴ R.(S.T)
R,S.T,(A.B).C ∴ (A.S).R
R ∴ R∨S
R ∴ S∨R
R.S ∴ S∨T
R.∼∼S ∴ R.(S∨T)
N.∼R,K.(F⊃H),(∼ U∨G).∼J ∴ ((F⊃H) .∼R).(∼ U∨G))
∼∼((F.G).(F.H)) ∴G.H
F,(F.∼∼G).(F.H)∴G∨H
F.G,F,∼∼H ∴ (G.F).(F.H)
F.(G.H) ∴ (F.G).H
(F.G).(F.H) ∴ F.(G.H)
F,G.H,(A.B).C ∴ (A.G).F
F.∼∼G ∴ F.(G∨H)
(F.H).(R.G) ∴G.FF⊃(G⊃H), F⊃G,F ∴H.F
F⊃((~G)⊃H), F⊃~G,F ∴H.F
F≡G, G ∴ H∨F
F≡G, G≡H, F ∴ H
F≡G, H.G ∴G.F(F⊃G), (G⊃H) ∴ (F⊃H)
(F⊃(G⊃H)), (F⊃G) ∴ (F⊃H)
(F⊃G) ∴ (F⊃(F.G))
∴ (F.G)⊃(G.F)
∴ ∼∼F⊃(∼F∨F)
∴ (F⊃(G⊃H))⊃((F.G)⊃H)
∴ ((F.G)⊃H)⊃(F⊃(G⊃H))∴ ((M.N).O)⊃(M.(N.O))
∴ (M⊃(N.O))⊃((M⊃N).(M⊃O))
F ∴ ∼(∼F)
F ∴ ∼(∼(∼(∼F)))
∼F ∴ ∼(F.G)
∴ F⊃∼∼F
F.∼G ∴ ∼(F⊃G)
F.∼G ∴ ∼(F≡G)
∴∼(F.(∼F))
C.(∼F), A.((∼G).B) ∴ ∼(F.G)R .(∼R) ∴ S
∼(R.S),R,S∴H
∴ R∨∼R
∴ R⊃∼∼R
∴ (R⊃S)⊃(∼S⊃∼R)
∴(∼S⊃∼R)⊃(R⊃S)
∼(R.S)∴∼R∨∼S
∴ ∼(R∨S)⊃(∼R.∼S)∴ (R⊃S)⊃(∼R∨S)
(F.F)∨(G.G),F⊃H,G⊃H∴H
F∨G ∴G∨F
(F⊃G).(H⊃D), F∨H ∴ G∨D
F∨(G.∼G)∴ F
F∨G, G⊃H ∴ ∼F⊃H
∼F.∼G∴ ∼(F∨G)
∴ (∼F.∼G)⊃∼(F∨G)
∴ (F.(G∨H))⊃((F.G)∨(F.H))∴ ((F∨G).(F∨H))⊃(F∨(G.H))
∴ (F.(G∨H))⊃((F.G)∨(F.H))
∴ ((F∨G).(F∨H))⊃(F∨(G.H))
∴ (∼F∨G)⊃(F⊃G)
∴ (F.F)≡F
∴ (F∨F)≡F
∴ F∨(G∨H)≡(F∨G)∨H
∴ F∨G≡G∨F∴ F∨G≡G∨F
∴ (∼F∨G)⊃(F⊃G)
∴ (∼F.∼G)⊃∼(F∨G)
∼F.∼G∴ ∼(F∨G)
F∨G, G⊃H ∴ ∼F⊃H
F∨(G.∼G)∴ F
(F⊃G).(H⊃D), F∨H ∴ G∨D
(F.F)∨(G.G),F⊃H,G⊃H∴H
∴ ((F.G)⊃H)⊃(F⊃(G⊃H))
(Z.W)⊃(L∨K),(W.Z)∴K∨ L
∼Y⊃∼Z,∼Z⊃∼X,∼X⊃∼Y∴Y≡Z
D⊃E,E⊃(Z.W),∼Z∨∼W ∴ ∼D
∴(A⊃B)∨(B⊃A)
∴((A⊃B)⊃A)⊃A)
∴(A≡ ∼A)⊃∼(A≡∼A)
∴A≡ ∼ ∼A
∴∼((A.B).∼(A.B))
F∨G,F⊃H,G⊃H∴H
(F.F)∨(G.G),F⊃H,G⊃H∴H
F∨G ∴G∨F
(F⊃G).(H⊃D), F∨H ∴ G∨D
F∨(G.∼G)∴ F
F∨G, G⊃H ∴ ∼F⊃H
∼F.∼G∴ ∼(F∨G)
∴ (F∨(F.G))≡F
∴ (F.(F∨G))≡F
∴ (F⊃(F⊃G))≡(F⊃G)
∴ F∨(G∨H)≡(F∨G)∨H
∴ F.(G.H)≡(F.G).H
∴ F∨G≡G∨F
∴ F.G≡G.F
∴ (F≡G)≡(G≡F)
∴ F⊃(G⊃H)≡G⊃(F⊃H)
∴(F⊃G)≡(∼G⊃∼F)
∴ ∼(F.G)≡∼F∨∼G
∴ ∼(F∨G)≡∼F.∼G
∴ F.(G∨H)≡(F.G)∨(F.H)
∴ F∨(G.H)≡(F∨G).(F∨H)
∴ F≡∼∼F
∴ (F≡G)≡(F⊃G).(G⊃F)
∴ (F≡G)≡(F.G)∨(∼F.∼G)
∴ F∨∼F
∴ (F⊃(G⊃H))≡((F.G)⊃H)
∴ (F.F)≡F
∴ (F∨F)≡F
∴ F⊃G≡∼F∨G
∴ (F⊃G)≡(∼F∨G)
∴ (F⊃G)≡∼(F.∼G)
∴ (F∨G)≡(∼F⊃G)
∴ (F.G)≡∼(F⊃∼G)
∴ (F≡G)≡ ((F⊃G) . (G⊃F))
F⊃G,G⊃H∴F⊃H
∴ F⊃G≡∼G⊃∼F