L2 inference rules | University of Edinburgh | PHIL08004 | 1 / 16 Alfred reads books ∴ Alfred reads books or God is an alien. 2 / 16 rule add Addition: from any sentence you may infer its disjunction with any other sentence. add: 2 (2 ∨ #) (# ∨ 2) 3 / 16 Either Alfred does yoga or Elle does yoga. But Alfred doesn’t do yoga ∴ Elle does yoga. 4 / 16 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 5 / 16 Alfred reads books and Alfred does yoga. ∴ Alfred does yoga. 6 / 16 rule s Simplification: if you have a conjunction, you may infer either conjunct. s: (2 ∧ #) 2 # 7 / 16 God is made of spaghetti. God owns a velociraptor. ∴ God is made of spaghetti and owns a velociraptor. 8 / 16 rule adj Adjunction: if you have any two sentences, you may infer their conjunction, in either order. adj: 2 # (2 ∧ #) (# ∧ 2) 9 / 16 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) 10 / 16 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 11 / 16 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? 12 / 16 13 / 16 rule bc Biconditional-to-conditional: from a biconditional you may infer either of the corresponding conditionals. bc: (2 ↔ #) (2 → #) (# → 2) 14 / 16 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 ↔ #) 15 / 16 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 16 / 16
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