if a certain natural deduction system is complete, then it is complete without material implication.

Consider 𝒲, the system of natural deduction articulated by the Howard-Snyders and Wasserman (2013). After introducing seventeen rules, some of which include Conditional Proof (CP), two forms of Reductio ad absurdum (RAA), Modus Ponens (MP), Simplification (Simp), Conjunction (Conj), Addition (Add), Commutation (Com), Double Negation (DN), and  Disjunctive Syllogism (DS), they introduce the rule of

    Material Implication (mi): 𝒫 ⊃ 𝒬  ∷ ¬𝒫 v 𝒬  

and they do so with the following justification:

Without material implication, our proof system would lack the capacity to prove valid every argument that is valid according to the truth table method. (p. 380)

One way to understand what the Howard-Snyders and Wasserman are sayings goes as follows: With (mi), 𝒲 can prove sentences of the form ¬𝒫 v 𝒬 from sentences of the form 𝒫 ⊃ 𝒬, and vice-versa. Or

    𝒫 ⊃ 𝒬 ⊢_𝒲 ¬𝒫 v 𝒬  

and

    ¬𝒫 v 𝒬 ⊢_𝒲  𝒫 ⊃ 𝒬

However, if we were to remove (mi) from 𝒲, perhaps forming the natural deduction system 𝒲*, we could not prove ¬𝒫 v 𝒬 from 𝒫 ⊃ 𝒬 and vice versa. That is:

    𝒫 ⊃ 𝒬 ⊬_𝒲*  ¬𝒫 v 𝒬

and

    ¬𝒫 v 𝒬 ⊬_𝒲*  𝒫 ⊃ 𝒬

We will now show that neither one is true—viz., we will show that

    𝒫 ⊃ 𝒬 ⊢_𝒲*  ¬𝒫 v 𝒬

and

    ¬𝒫 v 𝒬 ⊢_𝒲*  𝒫 ⊃ 𝒬

And we will do this by using only the inference rules of 𝒲*, which, as we’ve said is any one of 𝒲 but (mi). We shall begin with

    𝒫 ⊃ 𝒬 ⊢_𝒲*  ¬𝒫 v 𝒬

Consider a substitutional instance of 𝒫 ⊃ 𝒬:

    A ⊃ B

We shall show that

    A ⊃ B ⊢_𝒲*  ¬A v B

Proof:

1. A ⊃ B                                ⊢_𝒲*  ¬A v B

2.     ¬(¬A v B)                        Assume for RAA

3.         A                                 Assume for CP

4.         B                                 1, 3, MP

5.         B v ¬A                        4, Add

6.         ¬A v B                        5, Com

7.         ¬A v B • ¬(¬A v B)    6, 2, Conj

8.   ¬A v B                              2-7, RAA         

And now for

    ¬𝒫 v 𝒬 ⊢_𝒲*  𝒫 ⊃ 𝒬

Consider a substitutional instance of ¬𝒫 v 𝒬:

    ¬A v B

We shall show that

    ¬A v B ⊢_𝒲*  A ⊃ B

Proof:

1. ¬A v B      ⊢_𝒲*  A ⊃ B

2.     A             Assume for CP

3.     ¬¬A        2, DN

4.     B             1, 3, DS

5.  A ⊃ B       2-4, CP

What this shows is that, contra the Howard-Snyders and Wasserman, if their system of natural deduction is complete, viz. if it such that any sentence 𝒫 is a valid inference from a given set of premises, then there is a proof of 𝒫 from those same set of premises, it would still be complete without (mi).

References

Howard-Snyder, D. and F., and Wasserman, R. (2013). The Power of Logic (5^th Ed.) New York: McGraw Hill.

on William Lane craig on whether A logically implies B.

William Lane Craig (“Trinity Monotheism once more...”) says that,

Moreover, Howard-Snyder seems to assume that truth-making is closed under logical implication […] But that assumption is false. For example, “[a cat] has retractable claws” implies that “Grass is green,” since both are true, but they obviously have different truth-makers. (§3)

I have nothing to say about whether truth-making is closed under logical implication. I do have something to say about Craig’s argument that

    A    A cat has retractable claws. 

(logically) implies

    B    Grass is green.

“Since”, Craig says, “both are true”. Craig is almost certainly confused here.

By my lights, a sentence/statement/proposition 𝒫 logically implies 𝒬 , abbreviated 𝒫 ⊨ 𝒬, just in case 𝒬 is a logical consequence of 𝒫, which is to say that there is no truth-value assignment or “valuation” in which 𝒫 is true and 𝒬 is false. But, if you were to put A and B on a truth table, there will be one line in which A is true and B is false. Hence A does not logically imply B, or

    A ⊭ B

It may be, however, that all that Craig intended is to say is that, given that A and B are true, A “materially” implies B, or A ⊃ B. Indeed, since there is no truth value assignment in which A and B are true and A ⊃ B if false, the former logically implies the latter, or

    A & B ⊨ A ⊃ B

Note, however, that this doesn’t affect my previous point. It’s one thing to say that,

    given A & B, the material conditional A ⊃ B is true,

and it’s another to say that,

    given A & B, A logically implies B.

The former says something true about the conjunction, A & B, in relation to the material conditional, A ⊃ B, whereas the latter says something false about the conjunction, A & B, and its relation to another relation: the relation between A and B per se.

Now, by my understanding,

Relation ℛ is closed under logical implication just in case, if x stands in relation ℛ  to 𝒫, and if 𝒫 logically implies 𝒬, that is 𝒫 ⊨ 𝒬, then x stands in relation ℛ to 𝒬.

For example: some have thought, but many deny, that the knows that relation between an agent and a proposition is closed under logical implication. This view has the consequence that, if Craig knows that A, then Craig knows that: either A or ¬B, for A ⊨ A v ¬B. Notice that if the knows that relation were closed under logical implication, and we understood logical implication as “material” implication, then those who think that the knows that relation is closed under logical implication would be saddled with the view that, if Craig knows A, then Craig knows B, for A ⊃ B. As dubious as the view that the knows that relation is closed under logical implication might be, it's not that dubious.

Similarly, if we suppose that the makes true relation is closed under logical implication, then this would require only that, e.g., if x makes it true that A, then x makes it true that: either A or ¬B, for again A ⊨ A v ¬B. It would not require that, if x is makes it true that A, then x makes it true that B, as A ⊭ B, and this despite the fact that A and B are both true. In summary, then, it appears that Craig’s argument against the view that the makes true relation is closed under logical implication rests upon the following false conditional: If A and B, then A logically implies B.