2. Predicate Logic
● Similar to propositional logic for solving arguments, build from
quantifiers, predicates and logical connectives.
● A valid argument for predicate logic need not be a tautology.
● The meaning and the structure of the quantifiers and predicates
determines the interpretation and the validity of the arguments
● Basic approach to prove arguments:
■ Strip of quantifiers
■ Manipulate the unquantified wffs
■ Reinsert the quantifiers
Four new inference rules
Two rules to strip the qualifiers
Two rules to reinsert the qualifiers
● Note to remember: P(x) could be (∀y) (∀z) Q(x,y,z)
Section 1.4 Predicate Logic 1
3. Inference Rules
From
Can Derive
Name / Restrictions on Use
Abbreviation
(∀x)P(x)
P(t) where t is a Universal If t is a variable, it must
variable or constant Instantiation- ui
not fall within the scope
symbol
of a quantifier for t
(∃x)P(x)
P(t) where t is a Existential Must be the first rule
variable or constant Instantiation- ei
used that introduces t
symbol not previously
used in a proof
sequence
P(x)
(∀x)P(x)
Universal P(x) has not been
Generalization- ug
deduced from any
hypotheses in which x is
a free variable nor has P
(x) been deduced by ei
from any wff in which x
is a free variable
P(x) or P(a)
(∃x)P(x)
Existential To go from P(a) to (∃x)P
Generalization- eg
(x), x must not appear in
P(a)
Section 1.4 Predicate Logic 2
4. Examples: Proofs using Predicate Logic
● Prove the following argument:
■ All flowers are plants. Sunflower is a flower. Therefore, sunflower
is a plant.
■ P(x) is “ x is a plant”
■ a is a constant symbol (Sunflower)
■ Q(x) is “x is a flower”
● The argument is (∀x)[Q(x) → P(x)] Λ Q(a) → P(a)
● The proof sequence is as follows:
1. (∀x)[Q(x) → P(x)]
hyp
2. Q(a)
hyp
3. Q(a) → P(a)
1, ui
4. P(a)
2, 3, mp
Section 1.4 Predicate Logic 3
6. Examples: Proofs using Predicate Logic
● Prove the argument (∀x)[P(x) → Q(x)] Λ (∀x)P(x) → (∀x)Q(x)
● Proof sequence:
1. (∀x)[P(x) → Q(x)]
hyp
2. (∀x)P(x)
hyp
3. P(x) → Q(x)
1, ui
4. P(x)
2, ui : no restriction on ui about reusing a name
5. Q(x)
3, 4, mp
6. (∀x)Q(x)
5, ug
● Note: step 6 is legitimate since x is not a free variable in any
hypothesis nor was ei used before
Section 1.4 Predicate Logic 5
8. Examples: Proofs using Predicate Logic
● Prove the argument
(∀y)[P(x) → Q(x,y)] → [P(x) → (∀y)Q(x,y)]
● Using the deduction method, we can derive
(∀y)[P(x) → Q(x,y)] Λ P(x) → (∀y)Q(x,y)
● Proof sequence:
1. (∀y)[P(x) → Q(x,y)]
hyp
2. P(x)
hyp
3. P(x) → Q(x,y)
1, ui
4. Q(x,y)
2, 3, mp
5. (∀y)Q(x,y)
4, ug
Section 1.4 Predicate Logic 7
9. Temporary hypotheses
● A temporary hypothesis can be inserted into a proof sequence.
If T is inserted as a temporary hypothesis and eventually W is
deduced from T and other hypotheses, then the wff T → W has
been deduced from other hypotheses and can be reinserted into
the proof sequence
● Prove the argument
[P(x) → (∀y)Q(x,y)] → (∀y)[P(x) → Q(x,y)]
● Proof sequence:
1. P(x) → (∀y)Q(x,y)
hyp
2. P(x)
temporary hypothesis
3. (∀y)Q(x,y)
1, 2, mp
4. Q(x,y)
3, ui
5. P(x) → Q(x,y)
temp. hyp discharged
6. (∀y)[P(x) → Q(x,y)]
5, ug
Section 1.4 Predicate Logic 8
10. More Examples
● Prove the sequence [(∃x)A(x)]ʹ′ ⇔ (∀x)[A(x)]ʹ′
■ To prove equivalence, implication in each direction should be proved
● Proof sequence for [(∃x)A(x)]ʹ′ → (∀x)[A(x)]ʹ′
1. [(∃x)A(x)]ʹ′
hyp
2. A(x)
temp. hyp
3. (∃x)A(x)
2, eg
4. A(x) → (∃x)A(x)
temp. hyp discharged
5. [A(x)]ʹ′
1, 4, mt
6. (∀x)[A(x)]ʹ′
5, ug
● Proof sequence for (∀x)[A(x)]ʹ′ → [(∃x)A(x)]ʹ′
1. (∀x)[A(x)]ʹ′
hyp
2. (∃x)A(x)
temp. hyp
3. A(a)
2, ei
4. [A(a)]ʹ′
1, ui
5. [(∀x)[A(x)]ʹ′]ʹ′
3, 4, inc
6. (∃x)A(x) → [(∀x)[A(x)]ʹ′]ʹ′ temp. discharged
7. [((∀x)[A(x)]ʹ′)ʹ′]ʹ′ 1, dn
8. [(∃x)A(x)]ʹ′ 6, 7, mt
Section 1.4 Predicate Logic 9
11. Proving Verbal arguments
● Every crocodile is bigger than every alligator. Sam is a crocodile. But
there is a snake, and Sam isn’t bigger than that snake. Therefore,
something is not an alligator.
■ Use C(x): x is a crocodile; A(x): x is an alligator, B(x,y): x is bigger than y,
s is a constant (Sam), S(x): x is a Snake
● Hence prove argument
(∀x) (∀y)[C(x) Λ A(y) → B(x,y)] Λ C(s) Λ (∃x)(S(x) Λ [B(s,x)]ʹ′) → (∃x)[A(x)]
1. (∀x) (∀y)[C(x) Λ A(x) → B(x,y)] hyp
2. C(s) hyp
3. (∃x)(S(x) Λ [B(s,x)]ʹ′) hyp
4. (∀y)[C(s) Λ A(y) → B(s,y)] 1, ui
5. S(a) Λ [B(s,a)]ʹ′) 3, ei
6. C(s) Λ A(a) → B(s,a) 4, ui
7. [B(s,a)]ʹ′ 5, sim
8. [C(s) Λ A(a)]ʹ′ 6, 7, mt
9. [C(s)]ʹ′ V [A(a)]ʹ′ 9, De Morgan
10. [[C(s)]ʹ′]ʹ′ 2, dn
11. [A(a)]ʹ′ 9, 10, ds
12. (∃x)[A(x)]ʹ′ 11, eg
Section 1.4 Predicate Logic 10
12. Class Exercise
● Prove the argument
(∀x)[(B(x) V C(x)) → A(x)] → (∀x)[B(x) → A(x)]
Proof sequence:
1. (∀x)[(B(x) V C(x)) → A(x)] hyp
2. (B(x) V C(x)) → A(x) 1, ui
3. B(x) temp. hyp
4. B(x) V C(x) 3, add
5. A(x) 2, 4, mp
6. B(x) → A(x) temp. hyp discharged
7. (∀x)[B(x) → A(x)] 6, ug
Section 1.4 Predicate Logic 11
13. Class exercise
● Every ambassador speaks only to diplomats, and some ambassadors speak
to someone. Therefore, there is a diplomat.
● Use A(x): x is an ambassador; S(x,y): x speaks to y; D(x): x is a diplomat
Prove the argument
(∀x) (∀y)[(A(x) Λ S(x,y)) → D(y)] Λ (∃x)(∃y)(A(x) Λ S(x,y)) → (∃x)D(x)
Proof Sequence:
1. (∀x) (∀y)[(A(x) Λ S(x,y)) → D(y)] hyp
2. (∃x)(∃y)(A(x) Λ S(x,y)) hyp
3. (∀y)((A(a) Λ S(a,y)) → D(y)) 1, ui
4. (∃y)(A(a) Λ S(a,y)) 2, ei
5. A(a) Λ S(a,b) 4, ei
6. (A(a) Λ S(a,b)) → D(b) 3, ui
7. D(b) 5, 6, mp
8. (∃x)D(x) 7, eg
Section 1.4 Predicate Logic 12