Diagnosing proof fragments – using

**soundness**and**completeness**Proofs

A. Cut

B. Replace

Inferring types of sentences

**Soundness** says that if there is a proof of Γ ⊢ ϕ
then the argument from Γ to ϕ is valid (in the sense of truth
tables).

**Completeness** says that if the argument from Γ to ϕ
is valid then there is a proof of Γ ⊢ ϕ.

What these results mean in practice is that if the argument from Γ to ϕ is not valid, then there cannot be a line like this in a correctly written proof:

`Γ (n) ϕ`

Conversely, if Γ ⊢ ϕ is valid, then there *is* a correctly
written proof that ends with the line:

`Γ (n) ϕ`

Can there be a correctly written proof with line fragments like this?

```
1 (1) p∨q A
2 (2) p∨¬q A
1,2 (n) p
```

Yes: the sentences on lines 1 and 2 logically imply the sentence on line n. By completeness, there is a correctly written proof with these line fragments.

Suppose that we have two (correctly written) proofs as follows. What can we say about the semantic relationship between ϕ and ψ ?

```
1 (1) ϕ A
1 (17) ψ
```

```
1 (1) ψ A
1 (17) ϕ
```

The first proof and soundness show that ϕ logically implies ψ, and the second proof and soundness show that ψ logically implies ϕ. Therefore, ϕ and ψ are logically equivalent.

Can there be a correctly written proof with line fragments like this?

```
1 (1) ¬(p↔q)∧(¬(q↔r)∧¬(p↔r)) A
1 (17) p∧¬p
```

The formula on line 1 is an inconsistency. By completeness, there is
a proof of *any* sentence whatsoever depending on just 1.
Therefore, there is a correctly written proof with these line fragments.
(Isn’t it amazing that completeness guarantees that there always will be
a valid proof from any inconsistency to p∧¬p?)

Can there be a correctly written proof with line fragments like this?

```
1 (1) p∨¬p A
1 (17) q
```

No, the sentence on line 1 is a tautology while the sentence on line 17 is a contingency. So line 1 does not logically imply line 7. By soundness, there is no correctly written proof with these line fragments.

Can there be a correctly written proof with line fragments like this?

```
1 (1) p→(q∨r) A
1 (n) (p→q)∨r
```

What about this?

```
1 (1) (p→q)∨r A
1 (n) p→(q∨r)
```

Of course the answer to both questions is yes – as can be shown by a simple truth-table argument. Note that the only way for (p→q)∨r to be false is if p is true and q and r are false, which is also the only way for p→(q∨r) to be false. Hence, these two sentences are logically equivalent.

Suppose that ϕ and ψ are contingencies. Can we say anything definitive about the semantic type of ϕ∨ψ?

ϕ∨ψ cannot be an inconsistency, because ϕ is true on at least one row, and ϕ∨ψ is true whenever ϕ is true.

ϕ∨ψ could be a contingency, for example if ϕ is p and ψ is q.

ϕ∨ψ could be a tautology, for example if ϕ is p and ψ is ¬p.

Suppose that ϕ is a tautology and ψ is a contingency. Can we say anything definitive about the semantic type of ϕ→ψ?

In fact, when ϕ is a tautology, ϕ→ψ is logically equivalent to ψ. So it is a contingency if ψ is.

**Exercise:** Draw up a table where you check ϕ→ψ for
each possibility, where ϕ,ψ are tautology, inconsistency, or
contingency.

As Jillian pointed out, the following sequent is provable.

(*p*∧*q*) → *r* ⊢ (*p*→*r*) ∨ (*q*→*r*)

In fact, the meta-rules **cut** and
**replacement** can be used to give a fairly simple
proof.

```
1 (1) (p∧q)→r A
1 (2) ¬(p∧q)∨r 1 mat.con
1 (3) (¬p∨¬q)∨r 2 DM
1 (4) (¬p∨¬q)∨(r∨r) 3 duplication
1 (5) ¬p∨(¬q∨(r∨r)) 4 assoc
1 (6) ¬p∨((¬q∨r)∨r) 5 assoc
1 (7) ¬p∨(r∨(¬q∨r)) 6 comm
1 (8) (¬p∨r)∨(¬q∨r) 7 assoc
1 (9) (p→r)∨(¬q∨r) 8 mat.con
1 (10) (p→r)∨(q→r) 9 mat.con
```

**Exercise:** distinguish which moves could be ‘cut’ and
which moves have to be ‘replacement’. (Remember that cut only functions
on the entire formula on a line.)

**Exercise:** Give another proof of the above result
using a different strategy. For example, assuming (p∧q)→r, now assume
the negation of the conclusion and apply DeMorgans to get
¬(p→r)∧¬(q→r).

If we **substitute** p∧q for r in the above valid
sequent, we get the following valid sequent.

(*p*∧*q*) → (*p*∧*q*) ⊢ (*p*→(*p*∧*q*)) ∨ (*q*→(*p*∧*q*))

Since the premise here is a tautology, it follows that the conclusion is also a tautology! Think about that: it is guaranteed to be true that either p→(p∧q) or q→(p∧q). Does that seem right to you?

We can also check directly that this sentence is a tautology: if p→(p∧q) is false, then p is true and p∧q is false, which means that q is false; but then q→(p∧q) is true. Since one of the two disjuncts has to be true, the disjunction is always true.

**Advanced exercise:** Is (p→(p∧q))∨(q→(p∧q)) provable
in intuitionistic logic? (Hint: IL has the *disjunction
property*)