Lecture 5

Hans Halvorson

  1. Diagnosing proof fragments – using soundness and completeness

  2. Proofs

    A. Cut

    B. Replace

  3. Inferring types of sentences

Diagnosing proof fragments

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.

Inferring types of sentences

  1. 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. 

  2. 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)