Logic 1h: Simple applied arithmetic is logic
There’s exactly one F
There are exactly two G’s
Nothing is both an F and a G
Therefore, there are exactly three F’s and G’s (exactly three things that are either an F or a G)
I.e.,
∃x(Fx & ∀y(Fy → y=x)), ∃x∃y(((Gx & Gy) & x≠y) & ∀z(Gz → (z=x ∨ z=y))), ¬∃x(Fx & Gx)
∃x∃y∃z((((Fx∨Gx)&((Fy∨Fy)&(Fz∨Gz))) & (x≠y&(x≠z&y≠z))) & ∀w((Fw∨Gw)→(w=x∨(w=y∨w=z))))
There are no two ways about it: this is a long derivation.
1
2
3
4
5
6
4
4
6
6
6
6
6
6
15
4,15
4,6,15
4,6,15
3,4,6,15
3,4,6
21
4,21
4,6,21
4,6,21
3,4,6,21
3,4,6
6
28
29
4
4
4,29
4,29
34
6
6
6,34
6,34
4,6,28
4,6
4,6
3,4,6
3,4,6
6
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
(35)
(36)
(37)
(38)
(39)
(40)
(41)
(42)
(43)
(44)
∃x(Fx & ∀y(Fy → y=x))
∃x∃y(((Gx & Gy) & x≠y) & ∀z(Gz → (z=x ∨ z=y)))
¬∃x(Fx & Gx)
Fa & ∀y(Fy → y = a)
∃y(((Gb & Gy) & b≠y) & ∀z(Gz → (z=b ∨ z=y)))
((Gb & Gc) & b≠c) & ∀z(Gz → (z=b ∨ z=c))
Fa
Fa ∨ Ga
(Gb & Gc) & b ≠ c
Gb & Gc
Gb
Fb ∨ Gb
Gc
Fc ∨ Gc
a=b
Fb
Fb & Gb
∃x(Fx & Gx)
∃x(Fx & Gx) & ¬∃x(Fx & Gx)
a≠b
a=c
Fc
Fc & Gc
∃x(Fx & Gx)
∃x(Fx & Gx) & ¬∃x(Fx & Gx)
a≠c
b≠c
Fd ∨ Gd
Fd
∀y(Fy → y = a)
Fd → d = a
d=a
d = a ∨ (d = b ∨ d = c)
Gd
∀z(Gz → (z=b ∨ z=c))
Gd → (d = b ∨ d = c)
d=b∨d=c
d = a ∨ (d = b ∨ d = c)
d = a ∨ (d = b ∨ d = c)
(Fd ∨ Gd) → (d = a ∨ (d = b ∨ d = c))
∀w((Fw ∨ Gw) → (w=a ∨ (w=b ∨ w=c)))
a≠c&b≠c
a ≠ b & (a ≠ c & b ≠ c)
(Fb ∨ Gb) & (Fc ∨ Gc)
A
A
A
A
A
A
4 &-E
5 ∨-I
6 &-E
9 &-E
10 &-E
11 ∨-I
10 &-E
13 ∨-I
A (for RAA)
7, 15 =-E
11, 16 &-I
17 EI
3,18 &-I
15, 19 RAA
A (for RAA)
7, 21 =-E
13, 22 &-I
23 EI
3,24 &-I
21, 25 RAA
9 &-E
A (for CP)
A (for ∨-E)
4 &-E
30 UE
29, 31 MPP
32 ∨=I
A (for ∨-E)
6 &-E
35 UE
34,36 MPP
37 ∨-I [∨-E
28,29,33,34,38
28,39 CP
40 UI
26,27 &-I
20,42 &-I
12,14 &-I
4,6
3,4,6
3,4,6
(45)
(46)
(47)
3,4,6
(48)
3,4,5
(49)
3,4,5
(50)
2,3,4
(51)
2,3,4
(52)
1,2,3
(53)
(Fa ∨ Ga) & ((Fb ∨ Gb) & (Fc ∨ Gc))
8,44 &-I
((Fa∨Ga) & ((Fb∨Gb) & (Fc∨Gc))) & (a≠b & (a≠c & b≠c))
43,46 &-I
(((Fa∨Ga) & ((Fb∨Gb) & (Fc∨Gc))) & (a≠b & (a≠c & b≠c))) &
(∀w((Fw∨Gw) → (w=a ∨ (w=b ∨ w=c))))
41,46 &-I
∃z[(((Fa∨Ga) & ((Fb∨Gb) & (Fz∨Gz))) & (a≠b & (a≠z & b≠z))) &
(∀w((Fw∨Gw) → (w=a ∨ (w=b ∨ w=z))))]
47 EI
∃z[(((Fa∨Ga) & ((Fb∨Gb) & (Fz∨Gz))) & (a≠b & (a≠z & b≠z))) &
(∀w((Fw∨Gw) → (w=a ∨ (w=b ∨ w=z))))]
5,6,48 EE
∃y∃z[(((Fa∨Ga) & ((Fy∨Gy) & (Fz∨Gz))) & (a≠y & (a≠z & y≠z))) &
(∀w((Fw∨Gw) → (w=a ∨ (w=y ∨ w=z))))]
49 EI
∃y∃z[(((Fa∨Ga) & ((Fy∨Gy) & (Fz∨Gz))) & (a≠y & (a≠z & y≠z))) &
(∀w((Fw∨Gw) → (w=a ∨ (w=y ∨ w=z))))]
2,5,50 EE
∃x∃y∃z[(((Fx∨Gx) & ((Fy∨Gy) & (Fz∨Gz))) & (x≠y & (x≠z & y≠z))) &
(∀w((Fw∨Gw) → (w=x ∨ (w=y ∨ w=z))))]
51 EI
∃x∃y∃z[(((Fx∨Gx) & ((Fy∨Gy) & (Fz∨Gz))) & (x≠y & (x≠z & y≠z))) &
(∀w((Fw∨Gw) → (w=x ∨ (w=y ∨ w=z))))]
1,4,52 EE
With this example as guide we can work out that to show that
There are exactly m F’s
There are exactly n G’s
Nothing is both an F and a G
Therefore, there are exactly m+n F’s and G’s (exactly m+n things that are either an F or a G)
is valid as a matter of logic will take a derivation with somewhere approaching (7 × (m×n)) + (6 × (m+n)) + 25
lines. (The derivation above comes in at a little under that because ‘exactly one’ simplifies matters a little.)
Needless to say, multiplication is even more of a nightmare to simulate in logical terms but consider this:
There are (exactly) three cats in the basket.
Each cat in the basket is the mother of (exactly) two kittens.
No kitten has more than one mother. (Obviously!)
So there are six cats that are kittens of cats in the basket.
That’s valid.
Quite generally, inferences of this pattern are both valid and expressible using quantifiers and identity:
There are exactly n F’s.
Each F bears the relation R to exactly m things.
No two F’s bear the relation R to the same thing.
Therefore, there are exactly m×n things to which F’s bear the relation R.
Here’s what that looks like for m = n = 2:
∃x∃y(((Fx & Fy) & x≠y) & ∀z(Fz → (z=x ∨ z=y))),
∀x(Fx → ∃y∃z(((Rxy & Rxz) & x≠y) & ∀w(Rxw → (w=y ∨ w=z)))),
¬∃x∃y∃z(((Fx & Fy) & x≠y) & (Rxz & Ryz)),
∃x∃y∃z∃w[((∃v(Fv & Rvx) & (∃v(Fv & Rvy) & (∃v(Fv & Rvz) & ∃v(Fv & Rvw)))) &
(x≠y & (x≠z & (x≠w & (y≠z & (y≠w & z≠w)))))) &
∀u(∃v(Fv & Rvu) → (u=x ∨ (u=y ∨ (u=z ∨ u=w))))]
No, I’m not going to write out the derivation.