Deduction theorem

Let ‘P’ and ‘Q’ stand for (simple or compound) propositions.

The deduction theorem says that: if Q can be logically inferred from P, then ‘If P then Q’ can be proved as a theorem in the logical system in question.

This gives a method for dispensing with rules of inference in favor of axioms and theorems; but it does not hold for all logical systems, and in any case not all rules of inference can be dispensed with, for reasons due to Lewis Carroll.

L Carrol; What the Tortoise Said to Achilles

Examples of deduction

  1. “Prove” axiom 1:
    • P 1. hypothesis
    • Q 2. hypothesis
      • P 3. reiteration of 1
    • QP 4. deduction from 2 to 3
    P→(QP) 5. deduction from 1 to 4 QED
  2. “Prove” axiom 2:
    • P→(QR) 1. hypothesis
      • PQ 2. hypothesis
        • P 3. hypothesis
        • Q 4. modus ponens 3,2
        • QR 5. modus ponens 3,1
        • R 6. modus ponens 4,5
      • PR 7. deduction from 3 to 6
    • (PQ)→(PR) 8. deduction from 2 to 7
    (P→(QR))→((PQ)→(PR)) 9. deduction from 1 to 8 QED
  3. Using axiom 1 to show ((P→(QP))→R)→R:
    • (P→(QP))→R 1. hypothesis
    • P→(QP) 2. axiom 1
    • R 3. modus ponens 2,1
    ((P→(QP))→R)→R 4. deduction from 1 to 3 QED

Virtual rules of inference

From the examples, you can see that we have added three virtual (or extra and temporary) rules of inference to our normal axiomatic logic. These are “hypothesis”, “reiteration”, and “deduction”. The normal rules of inference (i.e. “modus ponens” and the various axioms) remain available.

1. Hypothesis is a step where one adds an additional premise to those already available. So, if your previous step S was deduced as:

{\displaystyle E_{1},E_{2},…,E_{n-1},E_{n}\vdash S,}

then one adds another premise H and gets:

{\displaystyle E_{1},E_{2},…,E_{n-1},E_{n},H\vdash H.}

This is symbolized by moving from the n-th level of indentation to the n+1-th level and saying

  • S previous step
    • H hypothesis

2. Reiteration is a step where one re-uses a previous step. In practice, this is only necessary when one wants to take a hypothesis which is not the most recent hypothesis and use it as the final step before a deduction step.

3. Deduction is a step where one removes the most recent hypothesis (still available) and prefixes it to the previous step. This is shown by unindenting one level as follows:

  • H hypothesis
  • ……… (other steps)
  • C (conclusion drawn from H)
  • HC deduction

Conversion from proof using the deduction meta-theorem to axiomatic proof

In axiomatic versions of propositional logic, one usually has among the axiom schemas (where PQ, and R are replaced by any propositions):

  • Axiom 1 is: P→(QP)
  • Axiom 2 is: (P→(QR))→((PQ)→(PR))
  • Modus ponens is: from P and PQ infer Q

These axiom schemas are chosen to enable one to derive the deduction theorem from them easily. So it might seem that we are begging the question. However, they can be justified by checking that they are tautologies using truth tables and that modus ponens preserves truth.

From these axiom schemas one can quickly deduce the theorem schema PP (reflexivity of implication) which is used below:

  1. (P→((QP)→P))→((P→(QP))→(PP)) from axiom schema 2 with P, (QP), P
  2. P→((QP)→P) from axiom schema 1 with P, (QP)
  3. (P→(QP))→(PP) from modus ponens applied to step 2 and step 1
  4. P→(QP) from axiom schema 1 with PQ
  5. PP from modus ponens applied to step 4 and step 3

Suppose that we have that Γ and H prove C, and we wish to show that Γ proves HC. For each step S in the deduction which is a premise in Γ (a reiteration step) or an axiom, we can apply modus ponens to the axiom 1, S→(HS), to get HS. If the step is H itself (a hypothesis step), we apply the theorem schema to get HH. If the step is the result of applying modus ponens to A and AS, we first make sure that these have been converted to HA and H→(AS) and then we take the axiom 2, (H→(AS))→((HA)→(HS)), and apply modus ponens to get (HA)→(HS) and then again to get HS. At the end of the proof we will have HC as required, except that now it only depends on Γ, not on H. So the deduction step will disappear, consolidated into the previous step which was the conclusion derived from H.

To minimize the complexity of the resulting proof, some preprocessing should be done before the conversion. Any steps (other than the conclusion) which do not actually depend on H should be moved up before the hypothesis step and unindented one level. And any other unnecessary steps (which are not used to get the conclusion or can be bypassed), such as reiterations which are not the conclusion, should be eliminated.

During the conversion, it may be useful to put all the applications of modus ponens to axiom 1 at the beginning of the deduction (right after the HH step).

When converting a modus ponens, if A is outside the scope of H, then it will be necessary to apply axiom 1, A→(HA), and modus ponens to get HA. Similarly, if AS is outside the scope of H, apply axiom 1, (AS)→(H→(AS)), and modus ponens to get H→(AS). It should not be necessary to do both of these, unless the modus ponens step is the conclusion, because if both are outside the scope, then the modus ponens should have been moved up before H and thus be outside the scope also.

Under the Curry–Howard correspondence, the above conversion process for the deduction meta-theorem is analogous to the conversion process from lambda calculus terms to terms of combinatory logic, where axiom 1 corresponds to the K combinator, and axiom 2 corresponds to the S combinator. Note that the I combinator corresponds to the theorem schema PP.

Helpful theorems

If one intends to convert a complicated proof using the deduction theorem to a straight-line proof not using the deduction theorem, then it would probably be useful to prove these theorems once and for all at the beginning and then use them to help with the conversion:

{\displaystyle A\to A}

helps convert the hypothesis steps.

{\displaystyle (B\to C)\to ((A\to B)\to (A\to C))}

helps convert modus ponens when the major premise is not dependent on the hypothesis, replaces axiom 2 while avoiding a use of axiom 1.

{\displaystyle (A\to (B\to C))\to (B\to (A\to C))}

helps convert modus ponens when the minor premise is not dependent on the hypothesis, replaces axiom 2 while avoiding a use of axiom 1.

These two theorems jointly can be used in lieu of axiom 2, although the converted proof would be more complicated:

{\displaystyle (A\to B)\to ((B\to C)\to (A\to C))}
{\displaystyle (A\to (A\to C))\to (A\to C)}

Peirce’s law is not a consequence of the deduction theorem, but it can be used with the deduction theorem to prove things which one might not otherwise be able to prove.

{\displaystyle ((A\to B)\to A)\to A}

It can also be used to get the second of the two theorems which can used in lieu of axiom 2.

Proof of the deduction theorem

We prove the deduction theorem in a Hilbert-style deductive system of propositional calculus.[4]

Let {\displaystyle \Delta } be a set of formulas and {\displaystyle A} and {\displaystyle B} formulas, such that {\displaystyle \Delta \cup \{A\}\vdash B}. We want to prove that {\displaystyle \Delta \vdash A\to B}.

Since {\displaystyle \Delta \cup \{A\}\vdash B}, there is a proof of {\displaystyle B} from {\displaystyle \Delta \cup \{A\}}. We prove the theorem by induction on the proof length n; thus the induction hypothesis is that for any {\displaystyle \Delta }{\displaystyle A} and {\displaystyle B} such that there is a proof of {\displaystyle B} from {\displaystyle \Delta \cup \{A\}} of length up to n{\displaystyle \Delta \vdash A\to B} holds.

If n = 1 then {\displaystyle B} is member of the set of formulas {\displaystyle \Delta \cup \{A\}}. Thus either {\displaystyle B=A}, in which case {\displaystyle A\to B} is simply {\displaystyle A\to A} which is derivable by substitution from p → p that is derivable from the axioms, hence also {\displaystyle \Delta \vdash A\to B}; or {\displaystyle B} is in {\displaystyle \Delta }, in which case {\displaystyle \Delta \vdash B}; it follows from axiom p → (q → p) with substitution that {\displaystyle \Delta \vdash B\to (A\to B)} and hence by modus ponens that {\displaystyle \Delta \vdash A\to B}.

Now let us assume the induction hypothesis for proofs of length up to n, and let {\displaystyle B} be a formula provable from {\displaystyle \Delta \cup \{A\}} with a proof of length n+1. Then there are three possibilities:

  1. {\displaystyle B} is member of the set of formulas {\displaystyle \Delta \cup \{A\}}; in this case we proceed as for n=0.
  2. {\displaystyle B} is arrived at by a substitution on a formula φ. Then φ is proven from {\displaystyle \Delta \cup \{A\}} with at most n steps, hence by the induction hypothesis {\displaystyle \Delta \vdash A\to \varphi }, where we may write A and φ with different variables. But then we may arrive from {\displaystyle A\to \varphi } at {\displaystyle A\to B} by the same substitution which is used to derive {\displaystyle B} from φ; thus {\displaystyle \Delta \vdash A\to B}.
  3. {\displaystyle B} is arrived at by using modus ponens. Then there is a formula C such that {\displaystyle \Delta \cup \{A\}} proves {\displaystyle C} and {\displaystyle C\to B}, and modus ponens is then used to prove {\displaystyle B}. The proofs of {\displaystyle C} and {\displaystyle C\to B} are with at most n steps, and by the induction hypothesis we have {\displaystyle \Delta \vdash A\to C} and {\displaystyle \Delta \vdash A\to (C\to B)}. By the axiom (p → (q → r)) → ((p → q) → (p → r)) with substitution it follows that {\displaystyle \Delta \vdash (A\to (C\to B))\to ((A\to C)\to (A\to B))}, and by using modus ponens twice we have {\displaystyle \Delta \vdash A\to B}.

Thus in all cases the theorem holds also for n+1, and by induction the deduction theorem is proven.

The deduction theorem in predicate logic

The deduction theorem is also valid in first-order logic in the following form:

  • If T is a theory and FG are formulas with F closed, and {\displaystyle T\cup \{F\}\vdash G}, then {\displaystyle T\vdash F\rightarrow G}.

Here, the symbol {\displaystyle \vdash } means “is a syntactical consequence of.” We indicate below how the proof of this deduction theorem differs from that of the deduction theorem in propositional calculus.

In the most common versions of the notion of formal proof, there are, in addition to the axiom schemes of propositional calculus (or the understanding that all tautologies of propositional calculus are to be taken as axiom schemes in their own right), quantifier axioms, and in addition to modus ponens, one additional rule of inference, known as the rule of generalization: “From K, infer ∀vK.”

In order to convert a proof of G from T∪{F} to one of FG from T, one deals with steps of the proof of G which are axioms or result from application of modus ponens in the same way as for proofs in propositional logic. Steps which result from application of the rule of generalization are dealt with via the following quantifier axiom (valid whenever the variable v is not free in formula H):

  • (∀v(HK))→(H→∀vK).

Since in our case F is assumed to be closed, we can take H to be F. This axiom allows one to deduce F→∀vK from FK and generalization, which is just what is needed whenever the rule of generalization is applied to some K in the proof of G.

In first-order logic, the restriction of that F be a closed formula can be relaxed given that the free variables in F has not been varied in the deduction of G from {\displaystyle T\cup \{F\}}. In the case that a free variable v in F has been varied in the deduction, we write {\displaystyle T\cup \{F\}\vdash ^{v}G} (the superscript in the turnstile indicating that v has been varied) and the corresponding form of the deduction theorem is {\displaystyle T\vdash (\forall vF)\rightarrow G}.[5]

Example of conversion

To illustrate how one can convert a natural deduction to the axiomatic form of proof, we apply it to the tautology Q→((QR)→R). In practice, it is usually enough to know that we could do this. We normally use the natural-deductive form in place of the much longer axiomatic proof.

First, we write a proof using a natural-deduction like method:

  • Q 1. hypothesis
    • QR 2. hypothesis
    • R 3. modus ponens 1,2
  • (QR)→R 4. deduction from 2 to 3
  • Q→((QR)→R) 5. deduction from 1 to 4 QED

Second, we convert the inner deduction to an axiomatic proof:

  • (QR)→(QR) 1. theorem schema (AA)
  • ((QR)→(QR))→(((QR)→Q)→((QR)→R)) 2. axiom 2
  • ((QR)→Q)→((QR)→R) 3. modus ponens 1,2
  • Q→((QR)→Q) 4. axiom 1
    • Q 5. hypothesis
    • (QR)→Q 6. modus ponens 5,4
    • (QR)→R 7. modus ponens 6,3
  • Q→((QR)→R) 8. deduction from 5 to 7 QED

Third, we convert the outer deduction to an axiomatic proof:

  • (QR)→(QR) 1. theorem schema (AA)
  • ((QR)→(QR))→(((QR)→Q)→((QR)→R)) 2. axiom 2
  • ((QR)→Q)→((QR)→R) 3. modus ponens 1,2
  • Q→((QR)→Q) 4. axiom 1
  • [((QR)→Q)→((QR)→R)]→[Q→(((QR)→Q)→((QR)→R))] 5. axiom 1
  • Q→(((QR)→Q)→((QR)→R)) 6. modus ponens 3,5
  • [Q→(((QR)→Q)→((QR)→R))]→([Q→((QR)→Q)]→[Q→((QR)→R))]) 7. axiom 2
  • [Q→((QR)→Q)]→[Q→((QR)→R))] 8. modus ponens 6,7
  • Q→((QR)→R)) 9. modus ponens 4,8 QED

These three steps can be stated succinctly using the Curry–Howard correspondence:

  • first, in lambda calculus, the function f = λa. λb. b a has type q → (q → r) → r
  • second, by lambda elimination on b, f = λa. s i (k a)
  • third, by lambda elimination on a, f = s (k (s i)) k

One thought on “Deduction theorem

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