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This homework assignment focuses on evaluating well-formed formulas (WFFs) in propositional logic. Given the evaluation of P, Q, and R as true, you are required to determine whether the provided WFFs are true or false based on logical rules. The WFFs provided include negations of biconditionals and conjunctions. The task involves consulting truth tables, analyzing the structure of the formulas, and confirming the evaluations. This exercise provides a strong foundation in understanding the principles of logical accuracy and the significance of parentheses in logical expressions.
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True or False? True or false? For each of the following WFFs, determine whether it is true or false on the given evaluation. Evaluation: P = T, Q = T, R = T 1. ~(P ↔ Q) 2. ~((P v (Q → R)) & ~P)
True or False? True or false? For each of the following WFFs, determine whether it is true or false on the given evaluation. Evaluation: P = T, Q = T, R = T 1. ~(P ↔ Q) 2. ~((P v (Q → R)) & ~P)
True or False? True or false? For each of the following WFFs, determine whether it is true or false on the given evaluation. Evaluation: P = T, Q = T, R = T 1. ~(P ↔ Q) 2. ~((P v (Q → R)) & ~P)
From My Inbox Dear Michael, Are there too many or too little parentheses in this formula: ~((P v (Q → R)) & ~P) Thanks, --Concerned Student
Nope! ~((P v (Q → R)) & ~P)
Set #1 ~((P v (Q → R)) & ~P)
Set #2 ~((P v(Q → R)) & ~P)
Set #3 ~((P v(Q → R))& ~P)
Definition of WFF • All sentence letters are WFFs. • If φ is a WFF, then ~φ is a WFF. • If φ and ψ are WFFs, then (φ & ψ), (φ v ψ), (φ → ψ), (φ ↔ ψ) are also WFFs. • Nothing else is a WFF.
Demonstration Using the definition we can show that certain sequences of symbols are WFFs. For example ~((P v (Q → R)) & ~P) is a WFF.
All Sentence Letters Are WFFs By (i), all sentence letters are WFFs. So: P is a WFF Q is a WFF R is a WFF
All Sentence Letters Are WFFs By (i), all sentence letters are WFFs. So: P is a WFF Q is a WFF R is a WFF
All Sentence Letters Are WFFs By (i), all sentence letters are WFFs. So: P is a WFF Q is a WFF R is a WFF
All Sentence Letters Are WFFs By (i), all sentence letters are WFFs. So: P is a WFF Q is a WFF R is a WFF
Scope Every occurrence of a connective in a WFF has a scope. The scope of that occurrence is the smallest WFF that contains it. For example The scope of “&” in “(~(~P&Q)→P)” is (~P&Q) • (~(~P & Q) → P) is not a WFF. • (~(~P & Q) → P)is not a WFF. • (~(~P & Q) → P) is a WFF, but is bigger than (~P&Q)
Occurrences Notice that the same symbol can occur different times in the same formula, and that its different occurrences can have different scopes. • ~((~P & Q) & (R ↔ Q)) • ~((~P & Q) & (R ↔ Q)) • ~((~P & Q) & (R ↔ Q)) • ~((~P & Q) & (R ↔ Q))
Scope of → ~((P v(Q → R))& ~P)
Scope of v ~((P v(Q → R))& ~P)
Scope of & ~((P v(Q → R))& ~P)
Scope of Main Connective ~((P v(Q → R))& ~P)