Lecture 3: Propositional Logic (III)
Fadoua Ghourabi
Ochanomizu University
[email protected]
Octobre 26, 2016
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Rules for negation: (9) Contradiction
There is no direct way to eliminate or introduce “¬”
Rules for negation involve the notion of
contradiction (⊥)
Contradictions are expressions of the form φ ∧ ¬φ
φ and ¬φ cannot be both true å Contradictory statements
(Result 1) All the contradictions are equivalent
Example.
p ∧ ¬p ` (r −→ s) ∧ ¬(r −→ s)
p ∧ ¬p a (r −→ s) ∧ ¬(r −→ s)
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Rules for negation: (9) Contradiction
There is no direct way to eliminate or introduce “¬”
Rules for negation involve the notion of
contradiction (⊥)
Contradictions are expressions of the form φ ∧ ¬φ
φ and ¬φ cannot be both true å Contradictory statements
(Result 1) All the contradictions are equivalent
Example.
p ∧ ¬p ` (r −→ s) ∧ ¬(r −→ s)
p ∧ ¬p a (r −→ s) ∧ ¬(r −→ s)
In other words:
p ∧ ¬p a` (r −→ s) ∧ ¬(r −→ s)
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Rules for negation: (9) Contradiction
There is no direct way to eliminate or introduce “¬”
Rules for negation involve the notion of
contradiction (⊥)
Contradictions are expressions of the form φ ∧ ¬φ
φ and ¬φ cannot be both true å Contradictory statements
(Result 1) All the contradictions are equivalent
Example.
p ∧ ¬p ` (r −→ s) ∧ ¬(r −→ s)
p ∧ ¬p a (r −→ s) ∧ ¬(r −→ s)
In other words:
p ∧ ¬p a` (r −→ s) ∧ ¬(r −→ s)
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Rules for negation:
(9) Contradiction & (10) not elimination
(Result 2) Any proposition can be derived from a
contradiction
φ ∧ ¬φ ` ψ
⊥
ψ
⊥e
¬φ
φ
¬e
⊥
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Rules for negation: Example
Prove that ¬p ∨ q ` p −→ q is valid
1.
2.
¬p ∨ q
premise
¬p
assumption
3.
p
assumption
4.
⊥
¬e 3,2
5.
q
⊥e 4
6.
p −→ q
⊥
ψ
⊥e
−→i 3-5
¬φ
φ
7.
q
assumption
8.
p
assumption
9.
q
copy 7
¬e
⊥
(Isabelle) False stands for ⊥
10.
11.
p −→ q
p −→ q
−→i 8-9
∨e 1, 2-6, 7-10
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Rules for negation: (11) not introduction
We assume φ and derive a contradiction ⊥.
Assumption φ cannot be true; so it must be
false, i.e. ¬φ.
φ.
..
.
⊥
¬φ
¬i
Prove that p −→ q, p −→ ¬q ` ¬p is valid
1.
p −→ q
premise
2.
p −→ ¬q
premise
3.
p
assumption
4.
q
mp 3, 1
5.
¬q
mp 3, 2
6.
⊥
¬e 4, 5
7.
¬p
¬i 3-6
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Rules for negation: (11) not introduction
We assume φ and derive a contradiction ⊥.
Assumption φ cannot be true; so it must be
false, i.e. ¬φ.
φ.
..
.
⊥
¬φ
¬i
Prove that p −→ q, p −→ ¬q ` ¬p is valid
1.
p −→ q
premise
2.
p −→ ¬q
premise
3.
p
assumption
4.
q
mp 3, 1
5.
¬q
mp 3, 2
6.
⊥
¬e 4, 5
7.
¬p
¬i 3-6
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Quiz 261016 (10 mins)
Back to our first example...
Prove that the reasoning below is correct.
Example 1.1
If the train arrives late and there are no taxis at the station, then John is
late for his meeting. John is not late for his meeting. The train did arrive
late. Therefore, there were taxis at the station.
Please send your .thy file to [email protected] (Today)
Good luck!
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