Download overhead 12/proofs in predicate logic [ov]

340
----->
example:
1. Fa
2. (x)(Fx  Gx)
3. (x)(Gx  Hx)
----->
Pr
Pr
Pr / (x)(Hx v Kx)
example:
1. (x)(Fx  Gx) Pr / Fb  (x)Gx
341
- UI, EG apply to WHOLE lines
1. (x)Fx  (x)Gx
2. Fa  Ga
UI 1
WRONG
1. (x)(Fx  Gx)
2. Fa  Ga
UI 2
OK
1. Fb  Gb
2. Fb  (x)Gx
EG 1
WRONG
1. Fb  Gb
2. (x)(Fx  Gx)
EG 1
OK
342
- REMEMBER to put in parens. to get scope right:
1. Fa
2. (x)(Fx  Gx)
3. (x)(Gx  Hx)
4. Fa  Ga
5. Ga  Ha
6. Fa  Ha
7. Ha
8. Ha v Ka
9. (x)Hx v Kx
Pr
Pr
Pr / (x)(Hx v Kx)
UI 2
UI 3
HS 4, 5
MP 1, 6
Add 7
EG 8
WRONG
- REMEMBER to put line numbers in
justifications:
1. Fa
Pr
2. (x)(Fx  Gx)
Pr
3. (x)(Gx  Hx) Pr / (x)(Hx v Kx)
4. Fa  Ga
UI
WRONG
5. Ga  Ha
UI
WRONG
- REMEMBER to put in individual variables in
propositional functions and individual constants
in instances:
1. Fa
Pr
2. (x)(Fx  Gx)
Pr
3. (x)(Gx  Hx) Pr / (x)(Hx v Kx)
4. F  G
UI 2
WRONG
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RECALL that each quantifier (that is, (x) and (x)) has
two rules:
- one for GETTING RID of the quantifier
- one for INTRODUCING the quantifier
and so far, we've talked about:
- the rule for getting rid of the universal
quantifier:
Universal Instantiation (UI)
(x)x
a
- the rule for introducing the existential
quantifier:
Existential Generalization (EG)
a
(x)x
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- note that:
- "instantiation" involves getting rid of an
quantifier by inferring an instance
- "generalization" involves introducing a
quantifier by generalizing from an instance
345
NOW the two remaining rules:
- the rule for getting rid of the existential quantifier:
Existential Instantiation (EI) (preliminary version)
(x)x
a
- the rule for introducing the universal quantifier:
Universal Generalization (UG) (preliminary
version)
a
(x)x
- these rules should seem much less intuitive than the
first two--in fact, they should seem illegitimate as
stated. In fact they are illegitimate as stated, so use of
them is restricted in important ways.
346
The rule for getting rid of the existential quantifier:
Existential Instantiation (EI) (preliminary version)
(x)x
a
Where x stands for any simple or compound
propositional function, and a stands for any
substitution instance of the propositional
function.
- the idea is: from an existential statement you can
infer an instance; this is ok, SO LONG AS the
individual constant in the instance hasn't
appeared in prior lines of the proof
347
- consider what goes wrong here (Nx  x is a number;
Ox  x is odd; Ex  x is even):
1. (x)(Nx  Ox)
2. (x)(Nx  Ex)
3. Na  Oa
4. Na  Ea
5. Oa
6. Na  Ea  Oa
7. (x)(Nx  Ex  Ox)
Pr
Pr
EI 1
EI 2
Simp 3
Conj 4, 5
EG 6 WHAT??
- in line 4, we've said of an individual
constant--that we've already said is an odd
number in line 3--that it's an even number
- to avoid this, in line 4 we can use an individual
constant that we haven't used on prior lines
1. (x)(Nx  Ox)
2. (x)(Nx  Ex)
3. Na  Oa
4. Nb  Eb
5. Oa
6. Nb  Eb  Oa
Pr
Pr
EI 1
EI 2
Simp 3
Conj 4, 5
- and the derivation of (x)(Nx  Ex  Ox) doesn't
work (as it SHOULDN'T) because you can't apply
EG to line 6 to get this existential statement
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Existential Instantiation (EI) (final version)
(x)x provided we flag a
a
- the flagging is just to help us identify use
of individual constants so that we can
apply the following restriction:
R1 A letter being flagged must be new to the proof,
that is, it may not appear, either in a formula or as
a letter being flagged, previous to the step in
which it gets flagged.
349
----->
example proof using EI
Existential Instantiation (EI)
(x)x provided we flag a
a
R1 A letter being flagged must be new to the proof,
that is, it may not appear, either in a formula or as
a letter being flagged, previous to the step in
which it gets flagged.
1. (x)(Cx  Lx)
2. (x)Cx
3. Ca
4. Ca  La
5. La
6. (x)Lx
Pr
Pr / (x)Lx
EI 2 (flag a)
UI 1
MP 3, 4
EG 5
- use EI BEFORE UI since UI is unrestricted--you can
use UI anywhere with no problem; if we had done:
1. (x)(Cx  Lx)
2. (x)Cx
3. Ca  La
4. Cb
5. ???
Pr
Pr / (x)Lx
UI 1
EI 2 (flag b)
- we wouldn't have been able to use MP on line 5
350
The rule for introducing the universal quantifier:
Universal Generalization (UG) (preliminary
version)
a
(x)x
Where x stands for any simple or compound
propositional function, and a stands for any
substitution instance of the propositional
function.
- the idea is: from an instance of a propositional
function you can infer a universal statement with
that propositional function; this is ok, SO LONG
AS this instance was derived from universal
statements
351
Universal Generalization (UG) (final version)
flag a



a
(x)x
- to ensure that the instance from which we infer a
universal statement WAS derived from universal
statements, we require that this instance is derived
within a "flagged subproof" and apply the following
restrictions:
R1 A letter being flagged must be new to the proof,
that is, it may not appear, either in a formula or as
a letter being flagged, previous to the step in
which it gets flagged.
R2 A flagged letter may not appear either in the
premises or in the conclusion of a proof.
352
----->
example proof using UG
Universal Generalization (UG)
flag a



a
(x)x
- FINALLY we can prove our first example
argument:
P1 All whales are mammals.
P2 All mammals are animals.
C All whales are animals.
353
1. (x)(Wx  Mx)
2. (x)(Mx  Ax)
3.
flag a
4.
Wa  Ma
5.
Ma  Aa
6.
Wa  Aa
7. (x)(Wx  Ax)
Pr
Pr / (x)(Wx  Ax)
FS (UG)
UI 1
UI 2
HS 4, 5
UG 6
- the first line of the flagged subproof is "flag a"
(or "flag b" or "flag c". . .); call this the flagging
step
- the justification for the first line of the subproof
is FS (UG) (for Flagging Step, Universal
Generalization)
- the last line of the subproof is the instance from
which we infer a universal statement
- the application of UG in the line after the
subproof discharges the flagging step
- all the terminology (scope, scope marker, etc.)
that applies to subproofs using CP and IP applies
here as well
354
NOTICE how use of the flagged subproof and
application of the restrictions:
R1 A letter being flagged must be new to the proof,
that is, it may not appear, either in a formula or as
a letter being flagged, previous to the step in
which it gets flagged.
R2 A flagged letter may not appear either in the
premises or in the conclusion of a proof.
PREVENTS us from using an instance derived from
an existential statement to infer a universal statement:
1. (x)Cx
Pr / (x)Cx
2.
flag a FS (UG)
3.
Ca
EI (flag a) WRONG: violates R1
1. (x)Cx
Pr / (x)Cx
2.
flag a FS (UG)
3.
Cb
EI (flag b)
- line 3. is ok, but it won't help us get (x)Cx--the
instance from which we infer a universal
statement has got to be the letter flagged in the
flagging step, line 2.
355
Proof strategies in predicate logic:
I.
For proofs where premises/conclusion include
quantifier statements
II. For proofs where premises/conclusion include
truth functional compounds of quantifier
statements
A. For proofs where premises/conclusion include
NEGATIONS of quantifier statements
B. For proofs where premises/conclusion include
conjunctions, disjunctions, conditionals, and
biconditionals of quantifier statements
356
Proof strategies in predicate logic:
I.
For proofs where premises/conclusion include
quantifier statements
Follow this general pattern:
- FS (UG) (if needed to get universal statement)
- EI and/or UI (to get rid of quantifiers in
premises)
- use EI before UI
- 20 sentential rules (to get instance you want to
generalize over)
- that's the 8 inference rules, 10 replacement
rules, and CP and IP
- EG or UG (to get a quantifier statement)
357
----->
example:
1. (x)(Fx  Gx)
2. (x)Fx
----->
Pr
Pr / (x)Gx
example:
1. (x)(Kx  Lx)
2. (x)((Kx  Lx)  Mx)
Pr
Pr / (x)(Kx  Mx)
358
Strategies for proofs in predicate logic (cont.)
II. For proofs where premises/conclusion include
truth functional compounds of quantifier
statements
A. For proofs where premises/conclusion include
NEGATIONS of quantifier statements
Follow this general pattern:
- QN or CQN rules (if needed to turn negated
quantifier statements in premises into quantifier
statements)
- FS (UG) (if needed to get universal statement)
- EI and/or UI (to get rid of quantifiers in premises
or results of QN or CQN rule applications)
- 20 sentential rules (to get instance you want to
generalize over)
- EG or UG (to get a quantifier statement)
- QN or CQN rules (if needed to turn quantifier
statement into a negated quantifier statement)
359
----->
example:
1. (x)(Fx  (Hx v Ix))
Pr
2. ~(x)(Fx  Hx)
Pr
3. (x)(Ix  (~Zx  Hx)) Pr / ~(x)(Fx  ~Zx)
360
----->
example:
1. ~(x)(Mx  Nx)
2. (x)(Px  Nx)
Pr
Pr / ~(x)(Mx  Px)
361
Strategies for proofs in predicate logic (cont.)
B. For proofs where premises/conclusion include
conjunctions, disjunctions, conditionals, and
biconditionals of quantifier statements
COMBINE:
- strategies you've learned to get
conjunctions, disjunctions,
conditionals, and biconditionals
- in particular, REMEMBER to use
CP to get conditionals
also, remember strategies such as:
- if conclusion is a conjunction,
you'll probably get the conclusion
by getting each conjunct on a line
by itself and using Conj
- if conclusion is a disjunction, try
to get one disjunct on a line by
itself and use Add
WITH:
- patterns that you've learned to get
quantifier statements and negations of
quantifier statements
362
----->
example:
1. (x)(Fx  Gx)
Pr / ~(x)Fx v (x)Gx
363
----->
example:
1. (x)Ax  (x)Bx
2. (x)Cx  (x)Dx
3. An  Cn
Pr
Pr
Pr / (x)(Bx  Dx)
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