60
Formal Logic
pRaCtiCe 22 Prove the argument
(4x)[P(x) S R(x)] ` [R( y)]′S [P( y)]′
Existential Instantiation
The existential instantiation rule allows us to strip off an existential quantifier.
It says that from (E x)P(x) we can derive P(a) or P(b) or P(c ) provided that these
are new constant symbols. The justification is that if P is true for some element
of the domain, we can give that element a specific name, but we cannot assume
anything else about it.
Without the restriction on universal instantiation, a hypothesis of the form
(4x)(E y)P(x, y) could lead to the wff (E y)P( y, y); here y has been substituted for x
within the scope of a quantifier on y. This would be invalid. For example, in the
domain of the integers, if P(x, y) means “y > x,” then (4x)(E y)P(x, y) is true (for
every integer there is a bigger integer) but (E y)P( y, y) is false (no integer has the
property that it is bigger than itself).
the argument is
(4x)[H(x) S M(x)] ` H(s) S M(s)
and a proof sequence is
(4x)(H(x) S M(x)) hyp
H(s)
hyp
H(s) S M(s)
1, ui
M(s)
2, 3, mp
In step 3, a constant symbol has been substituted for x throughout the scope of the
universal quantifier, as allowed by universal instantiation.
eXAMPLe 27
The following expressions would be legitimate steps in a proof sequence:
1. (4x)[P(x) S Q(x)] hyp
2. (E y)P( y)
hyp
3. P(a)
2, ei
4. P(a) S Q(a)
1, ui
5. Q(a)
3, 4, mp
In step 3, the specific element with property P was given the name a. In step 4,
ui was then used to say that an implication that is universally true in the domain
is certainly true for this a. Steps 3 and 4 cannot be reversed. If ui is first used on
hypothesis 1 to name a constant a, there is then no reason to assume that this particular a is the one that is guaranteed by hypothesis 2 to have property P.
Formal Logic
pRaCtiCe 22 Prove the argument
(4x)[P(x) S R(x)] ` [R( y)]′S [P( y)]′
Existential Instantiation
The existential instantiation rule allows us to strip off an existential quantifier.
It says that from (E x)P(x) we can derive P(a) or P(b) or P(c ) provided that these
are new constant symbols. The justification is that if P is true for some element
of the domain, we can give that element a specific name, but we cannot assume
anything else about it.
Without the restriction on universal instantiation, a hypothesis of the form
(4x)(E y)P(x, y) could lead to the wff (E y)P( y, y); here y has been substituted for x
within the scope of a quantifier on y. This would be invalid. For example, in the
domain of the integers, if P(x, y) means “y > x,” then (4x)(E y)P(x, y) is true (for
every integer there is a bigger integer) but (E y)P( y, y) is false (no integer has the
property that it is bigger than itself).
the argument is
(4x)[H(x) S M(x)] ` H(s) S M(s)
and a proof sequence is
(4x)(H(x) S M(x)) hyp
H(s)
hyp
H(s) S M(s)
1, ui
M(s)
2, 3, mp
In step 3, a constant symbol has been substituted for x throughout the scope of the
universal quantifier, as allowed by universal instantiation.
eXAMPLe 27
The following expressions would be legitimate steps in a proof sequence:
1. (4x)[P(x) S Q(x)] hyp
2. (E y)P( y)
hyp
3. P(a)
2, ei
4. P(a) S Q(a)
1, ui
5. Q(a)
3, 4, mp
In step 3, the specific element with property P was given the name a. In step 4,
ui was then used to say that an implication that is universally true in the domain
is certainly true for this a. Steps 3 and 4 cannot be reversed. If ui is first used on
hypothesis 1 to name a constant a, there is then no reason to assume that this particular a is the one that is guaranteed by hypothesis 2 to have property P.