Boolean logic gave us a collection of equivalences that allows us to
simplify (or complicate) expressions involving true/false variables
and logical relations. That’s all fine and good, but (our presentation
of) it was lacking any guidance about how to make an argument of the
form “if we know
Note that we will be adding the following logical relations.
T | T | T |
T | F | F |
F | T | T |
F | F | T |
T | T | T |
T | F | F |
F | T | F |
F | F | T |
Besides Boole’s and De Morgan’s rules in the Boolean logic notes, we
have some equivalences relating to
-
$x → y ≡ ¬ y → ¬ x$ (“contrapositive”) $x → y ≡ ¬ x ∨ y$ $x → (y → z) ≡ (x ∧ y) → z$ $x ↔ y ≡ (x → y) ∧ (y → x) ≡ (x ∧ y) ∨ (¬ x ∧ ¬ y)$
Suppose we have three sentences:
We can make an argument of the form:
$φ = x → y$ $ψ = ¬ y → x$ $ρ = y$
This argument is true if (and only if) there is no configuration of
truth values for
The idea is that
As an example, consider these assignments of the variables:
-
$φ =$ “All humans are mortal.” -
$ψ =$ “Socrates is a human.” -
$ρ =$ “Socrates is mortal.”
It’s not possible for
There are many patterns of logical consequences. Here are some that will be useful in our journey:
- Modus ponens (MP):
$φ = x → y$ $ψ = x$ $ρ = y$
- Modus tollens (MT):
$φ = x → y$ $ψ = ¬ y$ $ρ = ¬ x$
- Disjunctive syllogism (DS):
$φ = x ∨ y$ $ψ = ¬ x$ $ρ = y$
There is another form of this:
$φ = x ∨ y$ $ψ = ¬ y$ $ρ = x$
- Addition (Add):
$φ = x$ -
$ρ = x ∨ y$ (for any$y$ ; we have no need for$ψ$ here)
- Simplification (Simp):
$φ = x ∧ y$ -
$ρ = x$ (for any$y$ )
- Conjunction (Conj):
$φ = x$ $ψ = y$ $ρ = x ∧ y$
- Hypothetical syllogism (HS):
$φ = x → y$ $ψ = y → z$ $ρ = x → z$
- Constructive dilemma (CD):
$φ = (x → y) ∧ (w → z)$ $ψ = x ∨ w$ $ρ = y ∨ z$
- Absorption (Abs):
$φ = x → y$ $ρ = x → (x ∧ y)$
Now, given our relations and inference rules, we can collect some sentences and then deduce logical consequences of those sentences.
Suppose we know the following (don’t worry about what the variables actually mean; we’ll get to that issue later):
$c ∨ d$ $c → o$ $d → m$ $¬ o$
Ok, can we deduce that
-
$¬ c$ because of premises 2 and 4 plus modus tollens. -
$d$ because of premise 1 and consequence 5 and disjunctive syllogism. -
$m$ because of premise 3 and consequence 6 and modus ponens.