Proof by contradiction
(sub)goal
Theorem
Proof
If x2 is even, then x is even (x ∊ Z).
generating hypothesis
Let x∊ Z
pure hypothesis
Assume x!2 is even. Assume that x is odd.
conclusion
!
Then x = 2y+1 for some y ∊ Z.
2 = 4y2 + 4y + 1 =
Then x2 = if(2y+1)
y
is
not
free
in
P
and
2(2y2 + 2y) + 1 and 2y2 + 2y ∊ Z.
Q
!
So, x2 is odd
!
a contradiction.
So, x is even
Thanks to Bas Luttik
Proof by contradiction
How do we prove P by a contradiction?
¬P
proof by contradiction
(k)
{Assume}
¬P
…
val
val
F ⊨ ¬¬P ⊨
P
¬-intro
!
(l-1)
F
{¬-intro on (k) and (l-1)}
(l) ¬¬P
{¬¬-elim on (l)}
(l+1) P
(k < m)
¬¬-elim
time for an
example!
Disjunction introduction
How do we prove a disjunction?
val
¬P Q ⊨ P∨Q
!
val
¬Q P ⊨ P∨Q
∨-introduction
(k)
…
{Assume}
¬P
…
!
(l-1) Q
{∨-intro on (k) and (l-1)}
(l) P∨Q
-intro
Disjunction introduction
How do we prove a disjunction?
val
¬P Q ⊨ P∨Q
!
val
¬Q P ⊨ P∨Q
∨-introduction
(k)
…
{Assume}
¬Q
…
!
(l-1) P
{∨-intro on (k) and (l-1)}
(l) P∨Q
-intro
Disjunction elimination
How do we use a disjunction in a proof?
val
P ∨ Q ⊨ ¬P Q
!
val
P∨Q⊨
¬Q P
∨-elimination
|| ||
(k)
P∨Q
|| ||
!
{∨-elim on (k)}
(m) ¬P Q
(k < m)
Disjunction elimination
How do we use a disjunction in a proof?
val
P ∨ Q ⊨ ¬P Q
!
val
P∨Q⊨
¬Q P
∨-elimination
|| ||
(k)
P∨Q
|| ||
!
{∨-elim on (k)}
(m) ¬Q P
(k < m)
Proof by case distinction
How do we prove R by a case distinction?
proof by case distinction
!
(k)
!
(l)
!
|| ||
val
(P∨Q)⋀(P R)∧(Q R) ⊨ R
P∨Q
|| ||
P R
|| ||
(m) Q R
!
(n)
|| ||
{case-dist on (k), (l), (m)}
R
(k < n, l< n, m<n)
Bi-implication introduction
How do we prove a bi-implication?
-introduction
val
(P Q)∧(Q P) ⊨ P Q
…
(k)
P Q
…
!
(l)
Q P
∧-intro
…
{ -intro on (k) and (l)}
(m)
P Q
(k < m, l < m)
Bi-implication elimination
How do we use a bi-implication in a proof?
val
P Q ⊨ (P Q)∧(Q P)
-elimination
|| ||
(k)
|| ||
(k)
P Q
!
!
|| ||
|| ||
!
!
{ -elim on (k)}
(m)
P Q
P Q
{ -elim on (k)}
(m)
(k < m)
Q P
(k < m)
∧-elim
Derivations / Reasoning
with quantifiers
Proving a universal
quantification
To prove
Proof
∀x[x ∊ Z ∧ x ≥2 : x2 - 2x ≥0]
Let x∊ Z be arbitrary and assume that x≥2. !
Then, for this particular x, it holds that if y is
free in
x2 - 2x = x(x-2)
≥0not
(Why?)
!
Conclusion: ∀x[x ∊ Z ∧ x ≥2 : x2 - 2x ≥0].
P and Q
∀ introduction
How do we prove a universal quantification?
∀-introduction
(k)
…
{Assume}
var x; P(x)
…
similar to -intro
with
generating hypothesis
flag shows the validity of a
hypothesis
!
(l-1) Q(x)
{∀-intro on (k) and (l-1)}
(l) ∀x[P(x) : Q(x)]
Using a universal quantification
We know
∀x[x ∊ Z ∧ x ≥2 : x2 - 2x ≥0]
Whenever we encounter an a ∊ Z such that a≥2,
we can conclude that a2 - 2a ≥0.
!
For example, (523872 - 2·52387) ≥0 since 52387 ∊ Z and 52387 ≥2.
∀ elimination
How do we use a universal quantification in a proof?
∀-elimination
|| ||
(k)
∀x[P(x) : Q(x)]
|| ||
!
(l)
P(a)
similar to
implication but we need a witness
a is
an object (variable, number,..) which is “known” in line
(l)
!
time for an
example!
(m)
|| ||
{∀-elim on (k) and (l)}
Q(a)
(k < m, l < m)
the same “a” from line (l)
∃ introduction
How do we prove an existential quantification?
∃-introduction
(k)
val
¬ ∀x[P(x):¬Q(x)] ⊨
∃x [P(x) : Q(x)]
…
{Assume}
∀x[P(x) : ¬Q(x)]
…
!
(l-1) F
{∃-intro on (k) and (l-1)}
(l) ∃x [P(x) : Q(x)]
and ¬-intro
∃ elimination
How do we use an existential quantification in a proof?
val
∃x [P(x) : Q(x)] ⊨
¬ ∀x[P(x):¬Q(x)]
|| ||
∃-elimination
(k)
∃x [P(x) : Q(x)]
!
|| ||
!
(l)
∀x[P(x): ¬Q(x)]
and ¬elimination
!
(m)
|| ||
{∃-elim on (k) and (l)}
F
(k < m, l < m)
time for an
example!
Proofs with ∃-introduction and ∃elimination are unnecessarily long and
cumbersome…
There are alternatives!
Proving an existential
quantification
To prove
Proof
∃x[x ∊ Z : x3 - 2x - 8 ≥0]
It suffices to find a witness, i.e., an x∊ Z satisfying
x3 - 2x - 8 ≥0. !
x = 3 is a witness, since
andfree
33 - 2·3
= 13 ≥0
if y3is∊ Znot
in -P8 and
Q
!
Conclusion: ∃x[x ∊ Z : x3 - 2x - 8 ≥0].
also x = 5 is a witness…
Alternative ∃ introduction
How do we prove an existential quantification?
∃
by finding a witness
…
(k)
P(a)
…
!
(l)
(m)
Q(a)
…
{∃*-intro on (k) and (l)}
∃x [P(x) : Q(x)]
strategy: wait until a witness
object appears
(k < m, l < m)
does not
always work
Using an existential
quantification
We know
∃x[x ∊ R : a - x < 0 < b - x]
We can declare an x ∊ Z (a witness) such that
a - x < 0 < b - x
and use it further in the proof. For example:
From a - x < 0, we get a < x.
From b - x > 0, we get x < b.
Hence, a < b.
Alternative ∃ elimination
How do we use an existential quantification in a proof?
we pick a witness
∃
|| ||
(k)
∃x [P(x) : Q(x)]
!
x must be new!
|| ||
!
{∃*-elim on (k)}
(m) Pick x with P(x) and Q(x)
time for an
example!
(k < m)