{P} ⊦ Q if and only if {P} ╞ Q Meta theory and review Assume (as it is proven in Chapter 6 of our text) that: {P} ⊦ Q if and only if {P} ╞ Q The proof demonstrates that the system SD is sound or truth preserving: that you cannot, for example, derive a conclusion from a set of premises in SD unless the set of premises truth functionally entails the conclusion, and so forth.

{P} ⊦ Q if and only if {P} ╞ Q Meta theory and review And because: {P} ⊦ Q if and only if {P} ╞ Q Sentences P and Q are equivalent in SD IFF they are truth-functionally equivalent… And a set of sentences  of SL is inconsistent in SD iff  is truth-functionally inconsistent… And a sentence P is a theorem of SD IFF P is truth functionally true.

{P} ⊦ Q if and only if {P} ╞ Q Meta theory and review Explain why if {P} ⊦ Q if and only if {P} ╞ Q We would not want the following rule in SD: P v Q P ~Q

Meta theory and review Explain how and why the following rule in SD is truth preserving: P v Q P R Q R vE

Meta theory and review Explain why if a set {P, Q} is inconsistent in SD, any argument that has the set  as its premises will be valid in SD. a. What do we know if we know the set is inconsistent in SD? P Q R ~R

Meta theory and review a. What do we know if we know the set is inconsistent in SD? P Q R ~R ~S A R R ~R R S ~E

Review Strategies for constructing derivations in SD Check to see if the final sentence to be derived, or any sentence you need to get, is contained in the primary assumption or assumptions (if there are any) or former derived sentences in a way that an elimination rule will work. If so, identify the elimination rule you might use to get the sentence you need by identifying the main connective of the sentence in which the sentence you want to derive appears.

If you need P, see if P appears: Review If you need P, see if P appears: As part of a conjunction P & Q or Q & P so that you can use &E to get it. As the consequent of a material conditional Q  P so that if you can get the antecedent Q you can get P by  E. As one disjunct of a disjunction P v Q so that you might use vE to get it. As one half of a material biconditional P  Q so that if you can get the other half you can get P by E. As something you might get by assuming ~P and showing that a contradiction follows so as to get P by ~E.

Review If the final sentence to be derived (or any sentence you need to get there) is not contained in the primary assumption or assumptions (if there are any) or former derived sentences in a way that an elimination rule will work. If so, assume you need to build the sentence and identify the introduction rule you need to use by identifying the main connective of the sentence you want to derive.

Review If a sentence you want to derive (either the final line of the derivation or one you need to get to that final line) needs to be built and is of the form P & Q, identify the two conjuncts you need and add them to the derivation and then look to see how to get them. If that sentence is of the form P v Q, look to see which disjunct you might be able to get, write that in, and then work to get it and then P v Q by v I.

Review If it is of the form P  Q, you need to construct a subderivation with P as the assumption and Q as its last line. Sketch that out and see how to get Q under the assumption of P. If you do derive Q, move out to the scope line to the left with P  Q. If it is of the form P  Q, sketch in 2 subderivations: one with P as the assumption and Q as what you derive, and vice versa. Work to finish each subderivation to move out one scope line to the left with P  Q.

Review If it is of the form P , and if and only if there are negations available (possible sentences Q and ~Q) and there is no other way you can see to build P, try assuming ~P and try using ~E to derive P. If it is of the form ~P, and if and only if there are negations available (possible sentences Q and ~Q) and there is no other way you can see to build ~P, try assuming P and try using ~I to derive P.

Show that {(A  B) & (C  A), C v A} ⊦ B

Show that ‘A v A’ and ‘A &A’ are equivalent in SD