Prove by induction that any amount greater than 3 cents can be obtained using 2-cent and 5-cent coins.
Solution: Let P.n : (∃x,y:ℕ |: 2x + 5y = n). We show (informally) that P.n holds for all n > 3, using induction on n.
Base Case: n=4. Taking x' = 2 and y' = 0, we have 2x' + 5y' = 4, which shows that P.4 holds. More formally, we have
P.4
= < defn of P >
(∃x,y |: 2x + 5y = 4)
<== < (9.28), ∃-Introduction, P:=2x+5y=4, x:=x,y, E:=2,0 >
(2x + 5y = 4)[x,y := 2,0]
= < textual substitution >
2×2 + 5×0 = 4
= < arithmetic >
true
Thus, P.4 <== true, which by (3.58) and then (3.73) allows us to conclude P.4.
Here is an alternative proof that uses Range weakening/strengthening (9.25) instead of (9.28):
P.4
= < defn of P >
(∃x,y |: 2x + 5y = 4)
<== < (3.29), (9.25), P:=2x+5y=4, x:=x,y, E:=2,0, R:=true, Q:=x=2 ∧ y=0 >
(∃x,y | x=2 ∧ y=0 : 2x + 5y = 4)
= < (8.20) Nesting >
(∃x | x=2 : (∃y | y=0 : 2x + 5y = 4))
= < (8.14) One-point rule, arithmetic >
(∃x | x=2 : 2x = 4))
= < (8.14) One-point rule, arithmetic >
4 = 4
= < = is reflexive >
true
Inductive Case: Let n≥4. As an IH (i.e., induction hypothesis), assume P.4 ∧ P.5 ∧ ... ∧ P.n. It turns out (as frequently happens) that we can prove P(n+1) using (only) P.n as an assumption. Let x' and y' be natural numbers such that 2x' + 5y' = n. (From the assumption P.n, we conclude that x' and y' exist.)
Case y' > 0: (i.e., at least one 5-cent coin is in the collection of coins that makes n cents). We take x = x' + 3 and y = y' - 1. (That is, we add three 2-cent coins and remove one 5-cent coin.) We have
2x + 5y
= < x = x' + 3, y = y' - 1 >
2(x' + 3) + 5(y' - 1)
= < algebra >
2x' + 6 + 5y' - 5
= < arithmetic >
2x' + 5y' + 1
= < assumption 2x' + 5y' = n >
n+1
Thus, we have shown (assuming 2x' + 5y' = n, where x'≥0 ∧ y' > 0) that P(n+1) holds (i.e., that there exist x and y satisfying x≥0 ∧ y≥0 such that 2x + 5y = n+1).
Note that we would not have been justified in taking y = y' - 1 had we not been considering only the case in which y'>0. This explains the need for considering Case y'>0 (which we just did) and Case y'=0 (which we do below).
Case y' = 0: (i.e., no 5-cent coin is in the collection of coins that makes n cents). Substituting 0 for y' in 2x' + 5y' = n, we get 2x' = n. Since n≥4, it must be that x'≥2. We take x = x' - 2 and y = y' + 1. (That is, we remove two 2-cent coins and add one 5-cent coin.)
2x + 5y
= < x = x' - 2, y = y' + 1 >
2(x' - 2) + 5(y' + 1)
= < algebra >
2x' - 4 + 5y' + 5
= < arithmetic >
2x' + 5y' + 1
= < assumption 2x' + 5y' = n >
n+1
Let us repeat the inductive step with a little more formality. Recall that the goal of that step is to prove
for an arbitrary n≥4. By (3.82a) (taking p:= P.4 ∧ P.5 ∧ ... ∧ P.n, q:= P.n, and r:= P(n+1)), it suffices to prove P.n ==> P(n+1), which is what we shall do.
Recall that P.n : (∃x,y:ℕ |: 2x + 5y = n). By Metatheorem Witness (9.30), then, to prove P.n ⇒ P(n+1) it suffices to prove
Appealing to Case analysis (4.6), we consider two cases: y'>0 and y'=0.
Case y'>0: Assuming the antecedant (2x' + 5y' = n), we prove the consequent:
P(n+1)
= < defn of P >
(∃x,y:ℕ |: 2x + 5y = n+1)
<== < (9.28), ∃-introduction, P:=2x+5y = n+1, x:=x,y, E:=x'+3,y'-1 >
2(x'+3) + 5(y'-1) = n+1
= < algebra >
2x' + 6 + 5y' - 5 = n+1
= < arithmetic, assumption 2x' + 5y' = n >
n+1 = n+1
= < = is reflexive >
true
Note that, because y is of type ℕ (natural number), the instantiation y:=y'-1 in the application of (9.28) is justified only if y'>0, which is, indeed, the case we were considering.
Case y'=0: Assuming the antecedant (2x' + 5y' = n), we prove the consequent:
P(n+1)
= < defn of P >
(∃x,y:ℕ |: 2x + 5y = n+1)
<== < (9.28), ∃-introduction, P:=2x+5y = n+1, x:=x,y, E:=x'-2,y'+1 >
2(x'-2) + 5(y'+1) = n+1
= < algebra >
2x' - 4 + 5y' + 5 = n+1
= < arithmetic, assumption 2x' + 5y' = n >
n+1 = n+1
= < = is reflexive >
true
But wait! Is our instantiation x:=x'-2 in the application of (9.28) justified? To be so, we must have x'≥2, or else x would not be being instantiated to a natural number. Hence, to verify that our above proof is valid, we should show that the assumptions 2x' + 5y' = n (the assumed antecedant), n≥4 (we are proving P.n ==> P(n+1) for n≥4), and y'=0 (the case we are considering), implies x'≥2.
2x' + 5y' = n ∧ n ≥ 4 ∧ y'=0
==> < 3.84a, 3.76b >
2x' + 5*0 = n ∧ n ≥ 4
= < arithmetic >
2x' = n ∧ n ≥ 4
==> < = and ≥ are transitive >
2x' ≥ 4
= < algebra >
x' ≥ 2