– 15 –CHECKING VALIDITY WITH TREES IV

Paal Antonsenhttps://sites.google.com/site/paalantonsen/

antonsep@tcd.ie

PI2006 FORMAL LOGIC

Previously. . .

I Last lecture we introduced four new rules to our tree method:

(∀x)A

A(a/x)

for any name a

X∼(∀x)A

∼A(a/x)

where a is a new name.

X(∃x)A

A(a/x)

where a is a new name

∼(∃x)A

∼A(a/x)

for any name a

What happens when trees don’t close?

I So far we have considered examples where the trees close. But whathappens when a completed tree remains open?

I If the completed tree(s) for the initial list X, ∼A remains open then theargument 〈X, A〉 is classified as invalid (there is no proof for 〈X, A〉 ).

The tree for 〈X ,A〉 remains open iff there isn’t a proof for 〈X ,A〉 (X 0 A).

Intuitively, an open branch represents a counterexample to the argument〈X ,A〉. A counterexample is an evaluation where the premises are true,but the conclusion is not.

I In the case of propositional logic, we read off an evaluation that makesall the premises true and the conclusion false.

I In the case of predicate logic, we read off a (counter) model that makesall the premises true and the conclusion false.

What happens when trees don’t close?

I So far we have considered examples where the trees close. But whathappens when a completed tree remains open?

I If the completed tree(s) for the initial list X, ∼A remains open then theargument 〈X, A〉 is classified as invalid (there is no proof for 〈X, A〉 ).

The tree for 〈X ,A〉 remains open iff there isn’t a proof for 〈X ,A〉 (X 0 A).

Intuitively, an open branch represents a counterexample to the argument〈X ,A〉. A counterexample is an evaluation where the premises are true,but the conclusion is not.

I In the case of propositional logic, we read off an evaluation that makesall the premises true and the conclusion false.

I In the case of predicate logic, we read off a (counter) model that makesall the premises true and the conclusion false.

What happens when trees don’t close?

I So far we have considered examples where the trees close. But whathappens when a completed tree remains open?

I If the completed tree(s) for the initial list X, ∼A remains open then theargument 〈X, A〉 is classified as invalid (there is no proof for 〈X, A〉 ).

The tree for 〈X ,A〉 remains open iff there isn’t a proof for 〈X ,A〉 (X 0 A).

Intuitively, an open branch represents a counterexample to the argument〈X ,A〉. A counterexample is an evaluation where the premises are true,but the conclusion is not.

I In the case of propositional logic, we read off an evaluation that makesall the premises true and the conclusion false.

I In the case of predicate logic, we read off a (counter) model that makesall the premises true and the conclusion false.

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx)Qa∼Pa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx)Qa∼Pa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx)Qa∼Pa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

Pa ⊃ Qa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

Pa ⊃ Qa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

1) Has the appropriate been applied to every formula to which a rule canbe applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

1) Has the appropriate rule been applied to every formula to which a rulecan be applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

1) Has the appropriate rule been applied to every formula to which a rulecan be applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

Yes 1) Has the appropriate rule been applied to every formula to which a rulecan be applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

Yes 1) Has the appropriate rule been applied to every formula to which a rulecan be applied?

2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

D = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s check that we have completed the tree.

Yes 1) Has the appropriate rule been applied to every formula to which a rulecan be applied?

Yes 2) For each branch, is it the case that every universal formula has hadevery name in that branch substituted into it?

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1, 2}I(a) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

Yes We start with specifying a domain D, which contains an individual foreach name occuring in the branch.

Yes We then assign an interpretation of each name, so that it picks out amember of D, in this case we only have 1.

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 2I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

Yes We start with specifying a domain D, which contains an individual foreach name occuring in the branch.

Yes We then assign an interpretation of each name, so that it picks out amember of D, in this case we only have 1.

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 1I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

Yes We start with specifying a domain D, which contains an individual foreach name occuring in the branch.

Yes We then assign an interpretation of each name, so that it picks out amember of D, in this case we only have 1.

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 1I(P) = {1}I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

If Pa occur in that branch, make sure that I(a) ∈ I(P)If ∼Pa occur in that branch, make sure that I(a) /∈ I(P)

If Qa occur in that branch, make sure that I(a) ∈ I(Q)If ∼Qa occur in that branch, make sure that I(a) /∈ I(Q)

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 1I(P) = ∅I(Q) = {2}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

If Pa occur in that branch, make sure that I(a) ∈ I(P)If ∼Pa occur in that branch, make sure that I(a) /∈ I(P)

If Qa occur in that branch, make sure that I(a) ∈ I(Q)If ∼Qa occur in that branch, make sure that I(a) /∈ I(Q)

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 1I(P) = ∅I(Q) = {1}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from the left hand branch.

If Pa occur in that branch, make sure that I(a) ∈ I(P)If ∼Pa occur in that branch, make sure that I(a) /∈ I(P)

If Qa occur in that branch, make sure that I(a) ∈ I(Q)If ∼Qa occur in that branch, make sure that I(a) /∈ I(Q)

Open trees: Example I

(∀x)(Px ⊃ Qx), Qa; therefore Pa

M = {1, 2}D = {1}I(a) = 1I(P) = ∅I(Q) = {1}

(∀x)(Px ⊃ Qx) /aQa∼Pa

XPa ⊃ Qa

∼Pa Qa

Let’s now read off a counter modelM from left hand branch.

If we considerM, we can see that all the premises of the argument aretrue inM, but the conclusion is not. So the argument is invalid.

If Qa occur in that branch, make sure that I(a) ∈ I(Q)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

(∃x)Px(∃x)Qx

∼(∃x)(Px & Qx)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

(∃x)Px(∃x)Qx

∼(∃x)(Px & Qx)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

(∃x)Px(∃x)Qx

∼(∃x)(Px & Qx)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /a(∃x)Qx

∼(∃x)(Px & Qx)

Pa

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /a(∃x)Qx

∼(∃x)(Px & Qx)

Pa

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b∼(∃x)(Px & Qx)

Pa

Qb

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b∼(∃x)(Px & Qx)

Pa

Qb

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a

Pa

Qb

∼(Pa & Qa)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a

Pa

Qb

∼(Pa & Qa)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a

Pa

Qb

X∼(Pa & Qa)

∼Pa ∼Qa

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

∼(Pb & Qb)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

∼(Pb & Qb)

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Example II: (∃x)Px, (∃x)Qx ; therefore (∃x)(Px & Qx)

MD = {1, 2}I(a) = 1I(b) = 2I(P) = {1}I(Q) = {2}

All the premises of theargument are true inM,but the conclusion is not.So the argument isinvalid.

X(∃x)Px /aX(∃x)Qx /b

∼(∃x)(Px & Qx) /a, b

Pa

Qb

X∼(Pa & Qa)

∼Pa

⊗

∼Qa

X∼(Pb & Qb)

∼Pb ∼Qb

⊗

Multiple quantifiers

I In all the examples so far we have only considered cases of formulascontaining one quantifier. But, of course we can make formulas thathave multiple quantifiers occuring in them.

There is someone all philosophers admireJenny is a philosopherTherefore, Jenny admires someone

Jenny = ax is a philosopher = Pxx admires y = Rxy

Logical form: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

Multiple quantifiers

I In all the examples so far we have only considered cases of formulascontaining one quantifier. But, of course we can make formulas thathave multiple quantifiers occuring in them.

There is someone all philosophers admireJenny is a philosopherTherefore, Jenny admires someone

Jenny = ax is a philosopher = Pxx admires y = Rxy

Logical form: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

Multiple quantifiers

I In all the examples so far we have only considered cases of formulascontaining one quantifier. But, of course we can make formulas thathave multiple quantifiers occuring in them.

There is someone all philosophers admireJenny is a philosopherTherefore, Jenny admires someone

Jenny = ax is a philosopher = Pxx admires y = Rxy

Logical form: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

Multiple quantifiers

I In all the examples so far we have only considered cases of formulascontaining one quantifier. But, of course we can make formulas thathave multiple quantifiers occuring in them.

There is someone all philosophers admireJenny is a philosopherTherefore, Jenny admires someone

Jenny = ax is a philosopher = Pxx admires y = Rxy

Logical form: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

(∃x)(∀y )(Py ⊃ Ryx)Pa

∼(∃x)Rax

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

(∃x)(∀y )(Py ⊃ Ryx)Pa

∼(∃x)Rax

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb)

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb)

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb) /a

Pa ⊃ Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb) /a

Pa ⊃ Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb) /a

XPa ⊃ Rab

∼Pa Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb) /a

XPa ⊃ Rab

∼Pa

⊗

Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax

(∀y )(Py ⊃ Ryb) /a

XPa ⊃ Rab

∼Pa

⊗

Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax /b

(∀y )(Py ⊃ Ryb) /a

XPa ⊃ Rab

∼Pa

⊗

Rab

∼Rab

All branches close, so the argument is classified as valid.

Example III: (∃x)(∀y )(Py ⊃ Ryx), Pa; therefore (∃x)Rax

X(∃x)(∀y )(Py ⊃ Ryx) /bPa

∼(∃x)Rax /b

(∀y )(Py ⊃ Ryb) /a

XPa ⊃ Rab

∼Pa

⊗

Rab

∼Rab

⊗

All branches close, so the argument is classified as valid.

Tautologies, contradictions, and contingent formulas

I Just as we did for propositional logic, we can define these notions inpredicate logic:

A formula A is a tautology iff A is true in every model.

A formula A is a contradiction iff A is false in every model.

A formula A is contingent iff there is a model in which A is true and amodel in which A is false.

I Fortunately, we can also use the tree method to check whether someformula is a tautology, a contradiction and contingent.

Tautologies, contradictions, and contingent formulas

If the completed tree for ∼A close then A is a tautology.

If the completed tree for A close then A is a contradiction.

Example

X∼(∀x)(Px ∨ ∼Px) /a

X∼(Pa ∨ ∼Pa)

∼Pa∼∼Paa

⊗

(∀x)(Px ∨ ∼Px)is a tautology

Example

X(∃x)(Px & ∼Px) /a

XPa & ∼Pa

Pa∼Paa

⊗

(∃x)(Px & ∼Px)is a contradiction

Tautologies, contradictions, and contingent formulas

If the completed tree for ∼A close then A is a tautology.

If the completed tree for A close then A is a contradiction.

Example

X∼(∀x)(Px ∨ ∼Px) /a

X∼(Pa ∨ ∼Pa)

∼Pa∼∼Paa

⊗

(∀x)(Px ∨ ∼Px)is a tautology

Example

X(∃x)(Px & ∼Px) /a

XPa & ∼Pa

Pa∼Paa

⊗

(∃x)(Px & ∼Px)is a contradiction

Tautologies, contradictions, and contingent formulas

I We can also check whether some formula A is contingent.

We start by constructing two trees:(i) one tree with only A as the intial list.(ii) one tree with only ∼A as the initial list.

If the completed tree for A is open (A is not a contradiction)

and the completeted tree for ∼A is open (A is not a tautology)

then we can then read off from (i) a model in which A is true, and from (ii) amodel in which ∼A is true, i.e. one in which A is false.

Therefore, A is a contingent formula.

Tautologies, contradictions, and contingent formulas

I We can also check whether some formula A is contingent.

We start by constructing two trees:(i) one tree with only A as the intial list.(ii) one tree with only ∼A as the initial list.

If the completed tree for A is open (A is not a contradiction)

and the completeted tree for ∼A is open (A is not a tautology)

then we can then read off from (i) a model in which A is true, and from (ii) amodel in which ∼A is true, i.e. one in which A is false.

Therefore, A is a contingent formula.

Tautologies, contradictions, and contingent formulas

I We can also check whether some formula A is contingent.

We start by constructing two trees:(i) one tree with only A as the intial list.(ii) one tree with only ∼A as the initial list.

If the completed tree for A is open (A is not a contradiction)

and the completeted tree for ∼A is open (A is not a tautology)

then we can then read off from (i) a model in which A is true, and from (ii) amodel in which ∼A is true, i.e. one in which A is false.

Therefore, A is a contingent formula.

What will we learn in this course?

I In this course we’ve described two logical systems.

Logical systems A description of a logical system comes in three parts:

Grammar: A description of what counts as a formula.

Semantics: A definition of truth on an evaluation (or truth in a model);and derivatively validity and related concepts.

Proofs: A description of what counts as a proof.

The learning outcomes(A) learn how to translate sentences in natural language into the formal

languages of propositional and predicate logic;

(B) learn how to check the validity of arguments in propositional andpredicate logic using the tree method;

(C) learn how use formal methods for philosophical ends, focusing onvagueness, indeterminacy, existence and a puzzle about reasoning asexamples.