3. Let F be the formal system whose set of well-formed formulas (wff's) includes precisely those strings of the form xΔyΔz, where each of x, y, and z is a string composed of ∗'s and whose axioms and inference rules are as follows:
| Axiom | R1 | R2 |
|---|---|---|
A wff is, in other words, any string containing zero or more ∗'s and exactly two Δ's. The wff that includes no ∗'s is accepted, without proof, as a theorem.
Rule 1 says, informally, that if we take a theorem and insert one ∗ at the beginning and two at the end, what we get is another theorem.
Rule 2 says, informally, that if we take a theorem and remove one ∗ from both the left and right ends, and add one ∗ in the region lying between the two Δ's, what we get is another theorem.
As an example, here is a proof of ∗Δ∗∗Δ∗∗∗∗, which we can also write as ∗Δ∗2Δ∗4, using the convention that, for any natural number r, ∗r stands for a string of ∗'s of length r.
| (1) | ΔΔ | (axiom) |
|---|---|---|
| (2) | ∗ΔΔ∗2 | (R1[u,v,w := λ, λ, λ] applied to (1)) |
| (3) | Δ∗Δ∗ | (R2[u,v,w := λ, λ, ∗] applied to (2)) |
| (4) | ∗Δ∗Δ∗3 | (R1[u,v,w := λ,∗,∗] applied to (3)) |
| (5) | ∗2Δ∗Δ∗5 | (R1[u,v,w := ∗,∗,∗3] applied to (4)) |
| (6) | ∗Δ∗2Δ∗4 | (R2[u,v,w := ∗,∗,∗4] applied to (5)) |
Note that λ refers to the empty string.
(a) Show a proof of ∗Δ∗3Δ∗5
Solution: (sketch) Starting with the axiom (which is the only place one can start), apply R1 four times, and then apply R2 three times. (This assumes, of course, that, for each k, the wff appearing in line k+1 of the proof is the result of applying a rule to the wff appearing in line k.)
For that matter, the order in which the rules are applied does not matter, except that in no proof is it possible to have applied R2 more often than R1. (That's because in any proof in which the number of applications of R1 and R2 are the same, the last wff has no ∗'s preceding the first Δ, and thus R2 cannot be applied.)
(b) Argue for the following claim:
For every pair of natural numbers k and m, ∗kΔ∗mΔ∗2k+m is a theorem.
Suggested Approach: Provide informal instructions for writing a proof of ∗kΔ∗mΔ∗2k+m for arbitrarily chosen values of k and m.
Solution: Generalizing from the solution to (a), any proof in which R1 is applied k+m times and R2 is applied m times yields the theorem ∗kΔ∗mΔ∗2k+m.
(c) Argue for the following claim:
If ∗kΔ∗mΔ∗n is a theorem, then 2k+m = n.
Suggested approach: Explain why it must be that every line of every proof is such that the wff on that line satisfies the stated condition. For one-line proofs, this is easy. For multiple-line proofs, you need only argue that the wff on the last line satisfies the stated condition, under the assumption that the wff's on all previous lines do.
Solution: The only theorem that can result from a one-line proof is the axiom, ΔΔ, which is ∗kΔ∗mΔ∗2k+m, with k = m = 0.
Now suppose that we have a p-line proof, where p > 1, and the wff appearing on each of the first p−1 lines is of the required form. Now consider the wff on line p. It must be the result of applying either R1 or R2 to a wff on a previous line. By assumption, there exist k and m such that that previous wff is ∗kΔ∗mΔ∗2k+m.
If R1 were the rule applied, then the wff on line p must be ∗k'Δ∗m'Δ∗n', with k' = k+1, m' = m, and n' = 2k+m + 2. But 2k' + m' = 2(k+1) + m = 2k + m + 2 = n'. Hence, the wff on line p is of the correct form.
Now suppose that R2 were the rule applied. In this case, the wff on line p must be ∗k'Δ∗m'Δ∗n', with k' = k-1, m' = m+1, and n' = 2k+m − 1. But then 2k' + m' = 2(k-1) + m+1 = 2k + m − 1 = n'. Hence, the wff on line p is of the correct form.
What the above shows is that every wff that appears in any proof must be of the stated form.