Lecture 17 Handout 1:
All of the Rules
1
Introduction Rule for ∧
✤
Here’s introduction rule for ∧:
✤
✤
If you have ϕ on line ℓ1 and
you have ψ on line ℓ2, then
you may write ϕ∧ψ on any
subsequent line ℓ>ℓ₁, ℓ₂.
In terms of a picture, rule is:
✤
In the rule, it does not matter
whether ℓ1 or ℓ2 comes first. So
you could have also written
ψ∧ϕ on line ℓ>ℓ₁, ℓ₂.
✤
Here’s an example:
✤
1. p
2. q
3. r
4. q∧r
5. p∧(q∧r)
✤
✤
✤
✤
ℓ1.
ℓ2.
ℓ.
ϕ
ψ
ϕ∧ψ
✤
✤
(I∧,ℓ1,ℓ2)
✤
2
(assumption)
(assumption)
(assumption)
(I∧ 2,3)
(I∧ 1,4)
Elimination Rule for ∧
✤
Here’s elimination rule for ∧:
✤
Or the following:
✤
✤
✤
If you have ϕ∧ψ on line ℓ1,
then you may write ϕ on any
subsequent line ℓ>ℓ₁, and
likewise you may write ψ on
any subsequent line ℓ>ℓ₁.
✤
In terms of a picture, rule is:
✤
✤
ℓ1.
ℓ.
ϕ∧ψ
ϕ
Here’s an example:
✤
1. p∧(q∧r)
2. q∧r
3. r
✤
3
ϕ∧ψ
ψ
✤
✤
(E∧,ℓ1)
ℓ1.
ℓ.
(E∧,ℓ1)
(assumption)
(E∧,1)
(E∧,2)
Elimination Rule for ➝
✤
Here’s elimination rule for ➝:
✤
If you have ϕ on line ℓ1, and
you have ϕ➝ψ on line ℓ2,
then you may write ψ on any
subsequent line ℓ>ℓ₁, ℓ2.
✤
In the rule, it does not matter
whether ℓ1 or ℓ2 comes first.
✤
Here’s an example:
✤
1. (p→q)∧(p→r)
2. p
3. p→q
4. p→r
5. q
6. r
7. q∧r
✤
✤
In terms of a picture, rule is:
✤
✤
✤
✤
✤
ℓ1.
ℓ2.
ℓ.
ϕ➝ψ
ϕ
ψ
✤
✤
(E➝,ℓ1,ℓ2)
✤
4
(assumption)
(assumption)
(E∧ 1)
(E∧ 1)
(E➝2,3)
(E➝2,4)
(I∧ 5,6)
Introduction Rule for ➝
✤
So without further ado, here’s
the introduction rule for ➝:
✤
✤
Suppose that consecutive
lines ℓ1-ℓn constitute a proof
with premise ϕ and
conclusion ψ. Then one may
introduce ϕ→ ψ at any
subsequent line ℓ>ℓn, so long
as one brackets off lines lines
ℓ1-ℓn and never appeals to
them again.
✤
1.
p➝(q∧r)
(assumption)
2. p
3. (q∧r)
4. q
5. r
6. r∧q
(assumption)
(E➝, 1,2)
(E∧, 3)
(E∧, 3)
(I∧ 4,5)
7. p➝(r∧q)
(I➝ 2-6)
✤
✤
✤
✤
✤
✤
5
Here’s an example (picture
version of rule on next page):
Introduction Rule for ➝
✤
So without further ado, here’s
the introduction rule for ➝:
✤
Suppose that consecutive
lines ℓ1-ℓn constitute a proof
with premise ϕ and
conclusion ψ. Then one may
introduce ϕ→ ψ at any
subsequent line ℓ>ℓn, so long
as one brackets off lines lines
ℓ1-ℓn and never appeals to
them again.
✤
Here’s the picture version of the
rule:
✤
1.
✤
6
✤
ℓ1. ϕ
✤
ℓn. ψ
ℓ. ϕ ➝ ψ
(assumption)
(I➝ ℓ1-ℓn)
Introduction Rule for ∨
✤
Here’s introduction rule for ∨:
✤
✤
✤
If you have ϕ on line ℓ1 then
you may write ϕ∨ψ on any
subsequent line ℓ>ℓ₁, and
likewise you may write ψ∨ϕ
on any subsequent line ℓ>ℓ₁.
Here’s the picture of the rule:
✤
✤
Note that the rule does not
require that ψ appeared on any
previous line. In many ways,
this is what gives the ∨ rule its
strength.
ϕ
ϕ∨ψ
(I∨,ℓ1)
✤
Here’s an example:
✤
1. p
2. (p∨q)➝r
3. p∨q
4. r
✤
✤
✤
7
ℓ1.
ℓ.
(assumption)
(assumption)
(I∨, 1)
(E➝, 2,3)
Elimination Rule for ∨
✤
Here’s elimination rule for ∨:
✤
✤
If you have ϕ∨ψ on line ℓ1,
and you have ϕ→ξ on line ℓ2,
and you have ψ→ξ on line ℓ3,
then you may write ξ on any
subsequent line ℓ>ℓ1, ℓ2, ℓ3.
✤
Here’s the picture of the rule:
✤
ℓ1. ϕ∨ψ
ℓ2. ϕ→ξ
ℓ3. ψ→ξ
ℓ. ξ
✤
✤
✤
Again, the order in which ℓ1, ℓ2,
ℓ3, doesn’t matter. All that
matters is that all three of these
lines come before ℓ.
8
(E∨, ℓ1,ℓ2,ℓ3)
The Symbol ⊥
✤
To describe the introduction
and elimination rules for
negation, we need to introduce
a special symbol.
✤
This symbol is ⊥ and is called
falsum. Intuitively, it’s a special
symbol for a contradiction, for
something that’s always false.
✤
Initially it may seem strange to
have one symbol which stands
for all contradictions.
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✤
However, recall when we
studied validity that we said
that contradictions imply
anything. Hence, any two
contradictions are equivalent.
✤
So if we wanted our deductive
system to match up with our
judgements of validity, perhaps
it’s plausible that we have one
symbol ⊥ which stands for all
contradictions.
Elimination Rule for ¬
✤
So here’s the elimination rule
for ¬, first in prose:
✤
If you have ϕ on line ℓ1 and
you have ¬ϕ on line ℓ2, then
you may write ⊥ on any
subsequent line ℓ> ℓ1, ℓ2.
✤
In this rule, it doesn’t matter
whether the line with ϕ
comes before or after the line
with ¬ϕ.
✤
Example:
✤
✤
Here’s the picture:
✤
✤
✤
✤
✤
ℓ1. ϕ
ℓ2. ¬ϕ
ℓ. ⊥
✤
✤
(E¬, ℓ1, ℓ2).
10
1. p➝ ¬q
2. q
3. p
4. ¬q
5. ⊥
(assumption)
(assumption)
(assumption)
(E➝ 1, 3)
(I ¬ 2, 4)
Introduction Rule for ¬
✤
Here’s introduction rule for ¬,
first stated in prose:
✤
Suppose that consecutive
lines ℓ1-ℓn constitute a proof
with premise ϕ and
conclusion ⊥. Then one may
introduce ¬ϕ at any
subsequent line ℓ>ℓn, so long
as one brackets off lines lines
ℓ1-ℓn and never appeals to
them again.
✤
Here’s the picture version of the
rule:
✤
1.
✤
11
✤
ℓ1. ϕ
✤
ℓn. ⊥
ℓ. ¬ϕ
(assumption)
(I¬, ℓ1-ℓn)
Repeat Rule
✤
✤
This doesn’t come up so often,
but you will see it occasionally.
In a picture, the rule is simple:
✤
ℓ₁. ϕ
✤
ℓ. ϕ
In prose, the rule says:
✤
✤
✤
If you have ϕ on line ℓ1, then
you can write ϕ again on any
subsequent line ℓ>ℓ₁.
In applying this, stay out of
closed boxes. That is, don’t use
this rule to repeat things in
closed boxes outside of them.
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(repeat ℓ).
✤
Initially it’s unclear why one
would ever need/want to use
this rule.
✤
This rule is mostly useful as an
aid or helper to applying the
introduction rule for ➝.
EFSQ Rule
✤
✤
The acronym EFSQ stands for
the latin phrase “ex falso sequitur
quodlibet” which translates as
“from a contradiction, anything
follows”.
✤
In prose, this rule says:
✤
✤
If you have falsum ⊥ on any
line ℓ1, then you can write
any formula ψ on any
subsequent line ℓ>ℓ₁.
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In a picture, this rule looks like:
✤
ℓ₁. ⊥
✤
ℓ. ψ
(EFSQ ℓ₁).
In applying this rule, keep in
mind that instances of ⊥ that
appear in closed off boxes
above line ℓ are off limits: don’t
appeal later on to anything in
an already closed off box.
Double Negation Rule, aka ¬¬Rule
✤
The last rule we learn is the
double-negation rule, aka ¬¬
rule. In prose, it reads:
✤
✤
If you have ¬¬ϕ at line ℓ1,
then you can write ϕ at any
subsequent line ℓ> ℓ1.
Here’s the picture version:
✤
ℓ1. ¬¬ϕ
✤
ℓ. ϕ
(¬¬ ℓ1).
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✤
Here’s a simple example.
Recall we gave in an earlier
lecture a really long proof to
show that ¬¬¬p ⊢¬p.
✤
Well, there’s a super simple
proof using the ¬¬ rule:
✤
1. ¬¬¬p
(assumption)
✤
2. ¬p
(¬¬ 1).
Introduction Rule for Existential
Quantifier
✤
Here’s the introduction rule for
the existential quantifier:
✤
✤
✤
In the rule, the a is an
individual constant, like a name
for a person (like Alice). But of
course the rule holds for other
individual constants too, like
a,b,c. Likewise, it works for
other choices of variables.
✤
Simple Example:
If you have ϕ(a) on a line
then you may write ∃ x ϕ(x)
on any subsequent line.
In terms of a picture, the rule is:
✤
✤
ℓ1. ϕ(a)
ℓ. ∃ x ϕ(x)
✤
I∃, ℓ1
✤
15
1. Fb ∧ Gb
2. ∃ y (Fy ∧ Gy)
I∃, 1
Elimination Rule for Universal
Quantifier
✤
Here’s the elimination rule for
the universal quantifier:
✤
✤
If you have ∀ x ϕ(x) on a line
then you may write ϕ(a) on
any subsequent line.
In terms of a picture, the rule is:
✤
Again, the rule works for any
choice of constant (b, c as
opposed to a), and any choice of
variable (y,z as opposed to x).
✤
Simple Example:
✤
✤
✤
✤
ℓ1. ∀ x ϕ(x)
ℓ. ϕ(a)
E∀, ℓ1
16
1. ∀ y (Fy→Gy)
2. Fc → Gc
E∀, 1
Introduction Rule for Universal
Quantifier
✤
Here’s the introduction rule for
the universal quantifier:
✤
✤
If you have ϕ(a) on a line ℓ1
then you may write ∀ x ϕ(x) on
any subsequent line, provided
that (i) a doesn’t occur in ϕ(x),
(ii) a doesn’t occur at the top of
a box containing ℓ1.
In terms of a picture, the rule is:
✤
ℓ1. ϕ(a)
✤ ℓ.
∀ x ϕ(x)
I∀, ℓ1
17
✤
Here’s an example:
∀ x ∀ y Rxy⊢ ∀ y ∀ x Ryx
✤
1. ∀ x ∀ y Rxy
(assumption)
✤
2. ∀ y Rby
(E∀, 1)
✤
3. Rba
(E∀, 2)
✤
4. ∀ x Rbx
(I∀, 3)
✤
5. ∀ y ∀ x Ryx
(I∀ 4)
Elimination Rule for Existential
Quantifier
✤
Here’s it is:
✤
If you have ∃ x ϕ(x) on a line ℓ1 and
you have ϕ(a)→ψ on a line ℓ2 then
you may write ψ on any subsequent
line, provided that (i) a doesn’t occur
in ϕ(x), (ii) a doesn’t occur at the top
of a box containing ℓ1, and (iii) a doesn’t occur in ψ.
✤
Here’s an example:
∃ x Rxx ⊢ ∃ x ∃ y Rxy
✤
1. ∃ x Rxx
(assumption)
2. Raa
(assumption)
3. ∃ y Ray
(I∃, 2)
4. ∃ x ∃ y Rxy
(I∃, 3)
5. Raa → ∃ x ∃ y Rxy (I→ 2-4)
6. ∃ x ∃ y Rxy
(E∃ 1,5)
✤
✤
✤
✤
✤
In terms of a picture, the rule is:
✤
ℓ1. ∃ x ϕ(x)
✤
ℓ2. ϕ(a)→ψ
✤
ℓ. ψ
✤
E∃, ℓ1, ℓ2
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