| 9398 | ∧I: Given A and B, we may derive A∧B [Lemmon] |
| Full Idea: And-Introduction (&I): Given A and B, we may derive A∧B as conclusion. This depends on their previous assumptions. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9394 | MPP: Given A and A→B, we may derive B [Lemmon] |
| Full Idea: Modus Ponendo Ponens (MPP): Given A and A→B, we may derive B as a conclusion. B will rest on any assumptions that have been made. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9399 | ∧E: Given A∧B, we may derive either A or B separately [Lemmon] |
| Full Idea: And-Elimination (∧E): Given A∧B, we may derive either A or B separately. The conclusions will depend on the assumptions of the premiss. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9401 | ∨E: Derive C from A∨B, if C can be derived both from A and from B [Lemmon] |
| Full Idea: Or-Elimination (∨E): Given A∨B, we may derive C if it is proved from A as assumption and from B as assumption. This will also depend on prior assumptions. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9395 | MTT: Given ¬B and A→B, we derive ¬A [Lemmon] |
| Full Idea: Modus Tollendo Tollens (MTT): Given ¬B and A→B, we derive ¬A as a conclusion. ¬A depends on any assumptions that have been made | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9396 | DN: Given A, we may derive ¬¬A [Lemmon] |
| Full Idea: Double Negation (DN): Given A, we may derive ¬¬A as a conclusion, and vice versa. The conclusion depends on the assumptions of the premiss. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9393 | A: we may assume any proposition at any stage [Lemmon] |
| Full Idea: Assumptions (A): any proposition may be introduced at any stage of a proof. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9400 | ∨I: Given either A or B separately, we may derive A∨B [Lemmon] |
| Full Idea: Or-Introduction (∨I): Given either A or B separately, we may derive A∨B as conclusion. This depends on the assumption of the premisses. | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9402 | RAA: If assuming A will prove B∧¬B, then derive ¬A [Lemmon] |
| Full Idea: Reduction ad Absurdum (RAA): Given a proof of B∧¬B from A as assumption, we may derive ¬A as conclusion, depending on the remaining assumptions (if any). | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 9397 | CP: Given a proof of B from A as assumption, we may derive A→B [Lemmon] |
| Full Idea: Conditional Proof (CP): Given a proof of B from A as assumption, we may derive A→B as conclusion, on the remaining assumptions (if any). | |
| From: E.J. Lemmon (Beginning Logic [1965], 1.5) |
| 13500 | Conditional Proof: infer a conditional, if the consequent can be deduced from the antecedent [Hart,WD] |
| Full Idea: A 'conditional proof' licenses inferences to a conditional from a deduction of its consequent from its antecedent. | |
| From: William D. Hart (The Evolution of Logic [2010], 4) | |
| A reaction: That is, a proof can be enshrined in an arrow. |
| 14273 | Conditional Proof is only valid if we accept the truth-functional reading of 'if' [Edgington] |
| Full Idea: Conditional Proof seems sound: 'From X and Y, it follows that Z. So from X it follows that if Y,Z'. Yet for no reading of 'if' which is stronger that the truth-functional reading is CP valid, at least if we accept ¬(A&¬B);A; therefore B. | |
| From: Dorothy Edgington (Conditionals (Stanf) [2006], 2.2) | |
| A reaction: See the section of ideas on Conditionals (filed under 'Modality') for a fuller picture of this issue. Edgington offers it as one of the main arguments in favour of the truth-functional reading of 'if' (though she rejects that reading). |
| 10987 | Three traditional names of rules are 'Simplification', 'Addition' and 'Disjunctive Syllogism' [Read] |
| Full Idea: Three traditional names for rules are 'Simplification' (P from 'P and Q'), 'Addition' ('P or Q' from P), and 'Disjunctive Syllogism' (Q from 'P or Q' and 'not-P'). | |
| From: Stephen Read (Thinking About Logic [1995], Ch.2) |
| 13524 | Deduction Theorem: T∪{P}|-Q, then T|-(P→Q), which justifies Conditional Proof [Wolf,RS] |
| Full Idea: Deduction Theorem: If T ∪ {P} |- Q, then T |- (P → Q). This is the formal justification of the method of conditional proof (CPP). Its converse holds, and is essentially modus ponens. | |
| From: Robert S. Wolf (A Tour through Mathematical Logic [2005], 1.3) |