L2 inference rules .1in

L2 inference rules
| University of Edinburgh | PHIL08004 |
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Alfred reads books ∴ Alfred reads books or God is an alien.
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rule add
Addition: from any sentence you may infer its disjunction
with any other sentence.
add:
2
(2 ∨ #)
(# ∨ 2)
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Either Alfred does yoga or Elle does yoga. But Alfred doesn’t
do yoga ∴ Elle does yoga.
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rule mtp
Modus tollendo ponens: from a disjunction and the negation
of one of its disjuncts you may infer the other disjunct.
mtp:
(2 ∨ #)
˜#
2
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Alfred reads books and Alfred does yoga. ∴ Alfred does yoga.
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rule s
Simplification: if you have a conjunction, you may infer either
conjunct.
s:
(2 ∧ #)
2
#
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God is made of spaghetti. God owns a velociraptor.
∴ God is made of spaghetti and owns a velociraptor.
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rule adj
Adjunction: if you have any two sentences, you may infer
their conjunction, in either order.
adj:
2
#
(2 ∧ #)
(# ∧ 2)
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Biconditional
I
“James is angry if and only if Ann is happy”
I
‘if and only if’ is a connective used to combine two
propositions
I
I
P iff Q
I
I
A biconditional says that the truth of either proposition
requires the truth of the other, i.e., either both propositions
are true, or both are false
(P → Q) and (Q → P)
(P ↔ Q)
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iff
I
x is a rectangle iff:
I
I
I
I
x
x
x
x
necessary but not sufficient
is a polygon
necessary but not sufficient
has four sides
is a quadrilateral with right angles necessary and sufficient
is a square
sufficient but not necessary
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Puzzle. Two of the inhabitants—A and B—are standing together
under a tree. A makes the following statement: “I am a knave iff
B is a knave.”
What is B?
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rule bc
Biconditional-to-conditional: from a biconditional you may
infer either of the corresponding conditionals.
bc:
(2 ↔ #)
(2 → #)
(# → 2)
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rule cb
Conditionals-to-biconditional: from two conditionals where
the antecedent of one is the consequent of the other, and
vice versa, you may infer a biconditional containing the parts
of the conditionals.
cb:
(2 → #)
(# → 2)
(2 ↔ #)
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Practice exercises
2:1 (P∨R). (¬R∧T). (P∨(Q∧T))→S ∴ S
2:3 (P↔¬Q)→R ∴ (¬R∧P)→Q
2:4 ¬(¬R→(Q∧T)). ¬S∨R ∴ ¬S
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