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104 Cards in this Set
- Front
- Back
Quantificational Indeterminacy (set)
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iff neither {P} nor {~ P} has a closed truth-tree
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Closed Truth-Tree
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A truth-tree each of whose branches is closed
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Identity Elimination
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Conjunction Decomposition
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(∀y) (∀x) Lyx
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"Everyone likes everyone."
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Negated Negation Decomposition
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Existential Elimination
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(∀x) (∃y) Lxy
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"Everyone likes someone."
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State ‘(∀x)Lxm’ in terms of (∃x)
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∼ (∃x) ∼ Lxm
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Biconditional Introduction
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Closed Branch
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A branch containing both an atomic sentence and the negation of that sentence
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Equivalence in PD
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Sentences P and Q are equivalent in PD iff P is derivable in PD from {Q} and Q is derivable from {P}
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Sentence of PL
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Where all variables are metavariables:
A formula P of PL is a sentence of PL iff no occurrence of a variable in P is free (unquantified). |
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Negated Conditional Decomposition
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Universal Decomposition
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Implication
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Is ‘(∀z)(Haz ⊃ (∀z)(Fz ⊃ Gza))’ a formula of PL?
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No: quantifier appears more than once within its own scope
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Universal Introduction in PDE
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(∃y) (∀x) Lxy
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"Someone is liked by everyone."
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Negation Introduction
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Atomic formula of PL
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Where all variables are metavariables:
Every expression of PL that is either a sentence letter of PL or an n-place predicate of PL followed by n-individual terms of PL is an atomic formula. No connectives or quantifiers. |
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Quantificational Consistency (set)
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Set does not have a closed truth-tree
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Quantificational Validity (set)
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An argument of PL with finite premises is q-valid iff the set consisting of the premises and the negation of the conclusion is closed
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Hypothetical Syllogism
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Quantificational Entailment (set)
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A finite set Delta of PL q-entails P iff Delta U {~ P} has a closed truth-tree
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Completed Open Branch
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A finite branch on which each sentence is one of the following:
1. A literal (atomic or negated atomic sentence) 2. A decomposed compound sentence that is not a universally quantified sentence 3. A universally quantified sentence (\/x)P so that P(a/x) occurs on the branch for each constant a that occurs on the branch and at least one substitution instance P(a/x) occurs on the branch |
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Existential Introduction in PDE
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Quantifier of PL
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Where all variables are metavariables:
An expression of PL of the form (∀x) or (∃x). (∀x) = Universal quantifier. (∃x) = Existential quantifier. |
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Disjunction Decomposition
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(∃y) (∃x) Lyx
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"Someone likes someone."
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Exportation
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Universal Introduction
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Formula of PL
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Where all variables are metavariables:
1. Every atomic formula of PL is a formula of PL. 2. If P is a formula of PL, so is ∼ P. 3. If P and Q are formulas of PL, so are (P & Q), (P ∨ Q), (P ⊃ Q), and (P ≡ Q). 4. If P is a formula of PL that contains at least one occurrence of x and no x-quantifier, then (∀x)P and (∃x)P are both formulas of PL. 5. Nothing is a formula of PL unless it can be formed by repeated applications of clauses 1 to 4. |
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Expression of PL
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Any sequence of the vocabulary of PL
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Theorem in PD
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A sentence P of PL is a theorem of PD iff it is derivable from the empty set
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True on an interpretation
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where all variables are metavariables:
a sentence P of PL is true on an interpretation I iff every variable assignment d (for I) satisfies P on I |
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Negated Biconditional Decomposition
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Negation Elimination
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Form E
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Where all variables are metavariables:
(∀x) (P ⊃ ~Q) |
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Negated Universal Decomposition
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Open sentence of PL
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A forumla of PL that is not a sentence of PL.
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Equivalence
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Is ‘(∀y)(Hay ⊃ (Fy ⊃ Gya))’ a forumula of PL?
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Yes
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Existential Elimination
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Symbolize in pseudo-English: 'The Roman general who defeated Pompey invaded both Gaul and Germany.'
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There is exactly one thing that is a Roman general and defeated Pompey and that thing invaded both Gaul and Germany.
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Form O
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Where all variables are metavariables:
(∃x) (P & ~Q) |
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Expand for UD with constants a, b, and c: (∃x)(Fx ⊃ Gx)
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[(Fa ⊃ Ga) ∨ (Fb ⊃ Gb)] ∨ (Fc ⊃ Gc)
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Identity Decomposition
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Modus Tollens
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Open Branch
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A branch that is not closed
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Reiteration
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What is the contradictory of Form A (∀x)(P⊃Q)?
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Form O: (∃x)(P & ~Q)
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Negated Conjunction Decomposition
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Quantificational Falsity (truth-tree)
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iff set {P} has closed truth-tree
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Quantificational Equivalence
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Sentences meta-P and meta-Q of PL/E are quantificationally equivalent iff there is no interpretation where meta-P and meta-Q have different truth-values
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Quantifier Negation (QN)
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State ‘(∃x)Lxm’ in terms of (∀x)
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~ (∀x) ~ Lxm
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Identity Introduction
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Quantificational Entailment (|=)
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a set Γ of sentences of PL quantificationally entails a sentence meta-P of PL iff there is no interpretation where every member of Γ is true and meta-P is false
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Inconsistency in PD
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A set Gamma of sentences of PL is inconsistent in PD iff there is a sentence P where P and ~ P are derivable in PD from Gamma
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what are the only quantificational properties thatcan be proven directly?
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consistency and indeterminacy
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What is the contradictory of Form E (∀x)(P ⊃ ~Q)?
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Form I (∃x) (P&Q)
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Disjunction Elimination
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Quantificational Truth (truth-tree)
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iff the set {~ P} has a closed truth tree
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Quantificational Indeterminacy
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A sentence meta-P of PL/PLE is quantificationally indeterminate iff meta-P is neither quantificationally true nor quantificationally false
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Existential Decomposition
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Existential Elimination in PDE
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Quantificational consistency
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a set of sentences of PL is quantificationally consistent iff there is at least one interpretation on whch all members of the set are true
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Conditional Decomposition
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Where all variables are metavariables and P does not contain x,
(∀x)Ax ⊃ P is equivalent to: |
(∃x)(Ax ⊃ P)
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Quantificational validity
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An argument of PL/E with a finite number of premises is quantificationally valid iff there is no interpretation on which every premise is true and the conclusion is false
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Universal Decomposition in PLE
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Commutation
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Logical operator of PL
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An expression of PL that is either a quantifier or truth-functional connective
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Disjunctive Syllogism
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Quantificational Inconsistency
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a set of sentences of PL is quantificationally inconsistent iff there is no interpretation on which all members of the set are true
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Completed Truth-Tree
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A truth-tree each of whose branches either is closed or is a completed open branch
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Conjunction Elimination
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Open Truth-Tree
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A truth-tree that is not closed
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Form A
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Where all variables are metavariables:
(∀x) (P ⊃ Q) |
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Expand for UD containing constants a and c:
(∀x)(Gac ∨ Fx) |
(Gac ∨ Fa) & (Gac ∨ Fc)
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Negated Disjunction Decomposition
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Quantificational Equivalence (set)
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iff the set {~ (P<>Q)} has a closed truth-tree
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"Everybody loves somebody sometime"
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U.D.: people and times
Px: x is a person Tx: x is a time Lxyz: x loves y at z (\/x)(Px ⊃ (∃y)(∃z)[(Py & Tz) & Lxyz] |
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Vocabulary of PL
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Sentence letters
Predicates Individual terms (constants/variables) Truth-functional connectives Quantifier symbols Punctuation marks |
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Universal Elimination in PDE
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Quantificational Falsity
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A sentence meta-P of PL/PLE is quantificationally false iff meta-P is false on every interpretation
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False on an interpretation
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where all variables are metavariables:
a sentence P of PL is true on an interpretation I iff no variable assignment d (for I) satisfies P on I |
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Existential Introduction
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Disjunction Introduction
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Where all variables are metavariables and P does not contain x,
(∃x)Ax ⊃ P is equivalent to: |
(∀x)(Ax ⊃ P)
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Validity in PD
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An argument of PL is valid in PD iff the conclusion of the argument is derivable from the set consisting of the premises
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Derivability in PD
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Sentence P is derivable in PD from set Delta iff there is a derivation in PD in which all of the primary assumptions are members of set Delta and P appears within the scope of only the primary assumptions
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Biconditional Decomposition
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Negated Existential Decomposition
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Quantificational truth
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A sentence meta-P of PL/PLE is quantificationally true iff meta-P is true on every interpretation
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Form I
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Where all variables are metavariables:
(∃x) (P & Q) |
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Universal Elimination
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Distribution
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Is ‘(∀x)(Haz ⊃ (∀z)(Fz ⊃ Gza))’ a forumla of PL?
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No: variable x does not appear within scope of x-quantifier
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Conjunction Introduction
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Quantificational Inconsistency (set)
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Set has a closed truth-tree
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Expand for UD containing constants a and b:
(∀x)(Wx ⊃ (∃y)Cxy) |
(Wa ⊃ (Caa ∨ Cab)) & (Wb ⊃ (Cba ∨ Cbb))
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Biconditional Elimination
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