First-Order Logic. Natural Deduction, Part 2
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1 First-Order Logic Natural Deduction, Part 2
2 Review Let ϕ be a formula, let x be an arbitrary variable, and let c be an arbitrary constant. ϕ[x/c] denotes the result of replacing all free occurrences of x in ϕ with c.
3 Review (Gzz Hxz) ( x)(sx Px) (Fy ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
4 Review (Gzz Hxz)[z/d] ( x)(sx Px) (Fy ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
5 Review (Gzz Hxz)[z/d] (Gdd Hxd) ( x)(sx Px) (Fy ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
6 Review (Gzz Hxz)[z/d] (Gdd Hxd) ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
7 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Gdd Hxd) ( x)(sx Px) (Fy ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
8 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Gdd Hxd) ( x)(sx Px) (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Rxx ᴧ ( x)fx) (~Dy Kz)
9 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Gdd Hxd) ( x)(sx Px) (Fc ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx) (~Dy Kz)
10 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Gdd Hxd) ( x)(sx Px) (Fc ᴧ ( y)(by ᴧ Dx)) (Rxx ᴧ ( x)fx)[x/b] (~Dy Kz)
11 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Rxx ᴧ ( x)fx)[x/b] (Gdd Hxd) ( x)(sx Px) (Fc ᴧ ( y)(by ᴧ Dx)) (Rbb ᴧ ( x)fx) (~Dy Kz)
12 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Rxx ᴧ ( x)fx)[x/b] (Gdd Hxd) ( x)(sx Px) (Fc ᴧ ( y)(by ᴧ Dx)) (Rbb ᴧ ( x)fx) (~Dy Kz)[y/a]
13 Review (Gzz Hxz)[z/d] ( x)(sx Px)[y/e] (Fy ᴧ ( y)(by ᴧ Dx))[y/c] (Rxx ᴧ ( x)fx)[x/b] (~Dy Kz)[y/a] (Gdd Hxd) ( x)(sx Px) (Fc ᴧ ( y)(by ᴧ Dx)) (Rbb ᴧ ( x)fx) (~Da Kz)
14 Universal Elimination ( E) Suppose ( x)ϕ is a well-formed formula for some variable x, and suppose c is an arbitrary constant. ( x)ϕ ϕ[x/c]
15 Let s try an example: { ( x)(gx ᴧ Hx), (Ga ( y)ky) } (Ha ᴧ Kb)
16 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A
17 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A 1 (3) (Ga ᴧ Ha) 1 E
18 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A 1 (3) (Ga ᴧ Ha) 1 E 1 (4) Ga 3 ᴧE 1 (5) Ha 3 ᴧE
19 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A 1 (3) (Ga ᴧ Ha) 1 E 1 (4) Ga 3 ᴧE 1 (5) Ha 3 ᴧE 1,2 (6) ( y)ky 2,4 E
20 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A 1 (3) (Ga ᴧ Ha) 1 E 1 (4) Ga 3 ᴧE 1 (5) Ha 3 ᴧE 1,2 (6) ( y)ky 2,4 E 1,2 (7) Kb 6 E
21 1 (1) ( x)(gx ᴧ Hx) A 2 (2) (Ga ( y)ky) A 1 (3) (Ga ᴧ Ha) 1 E 1 (4) Ga 3 ᴧE 1 (5) Ha 3 ᴧE 1,2 (6) ( y)ky 2,4 E 1,2 (7) Kb 6 E 1,2 (8) (Ha ᴧ Kb) 5,7 ᴧI
22 And here s another: { ( x)( y)rxy } Raa
23 1 (1) ( x)( y)rxy A
24 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E
25 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E 1 (3) Raa 2 E
26 Universal Introduction ( I) Suppose ϕ is a well-formed formula, and c is a constant that does not appear in ϕ. Then ϕ[x/c] ( x)ϕ So long as ϕ[x/c] does not depend on any formula containing c.
27 The rule for universal introduction has two clauses that constrain how we can introduce universal quantifiers. The constant c cannot appear in ϕ. The formula ϕ[x/c] cannot depend on any formula containing c.
28 The easiest way to see how the constraints on universal introduction work is to see how they might be violated.
29 The easiest way to see how the constraints on universal introduction work is to see how they might be violated. Let s see a proof that violates the first constraint: The constant c cannot appear in ϕ.
30 1 (1) ( x)fxx A
31 1 (1) ( x)fxx A 1 (2) Fbb 1 E
32 1 (1) ( x)fxx A 1 (2) Fbb 1 E 1 (3) ( x)fxb 2 I
33 1 (1) ( x)fxx A 1 (2) Fbb 1 E 1 (3) ( x)fxb 2 I NO!
34 1 (1) ( x)fxx A 1 (2) Fbb 1 E 1 (3) ( x)fxb 2 I NO! The formula Fxb contains the constant b!
35 1 (1) ( x)fxx A 1 (2) Fbb 1 E 1 (3) ( x)fxb 2 I NO! The formula Fxb contains the constant b! Construct a small world showing invalidity.
36 And now a proof that violates the second constraint: The formula ϕ[x/c] cannot depend on any formula containing c.
37 1 (1) Fa A
38 1 (1) Fa A 1 (2) ( x)fx 1 I
39 1 (1) Fa A 1 (2) ( x)fx 1 I NO!
40 1 (1) Fa A 1 (2) ( x)fx 1 I NO! The formula Fa depends on a formula that contains the constant a, namely itself!
41 1 (1) Fa A 1 (2) ( x)fx 1 I NO! The formula Fa depends on a formula that contains the constant a, namely itself! Construct a small world showing invalidity.
42 Now, let s see a correct example: { ( x)(fa Gx) } (Fa ( x)gx)
43 1 (1) ( x)(fa Gx) A
44 1 (1) ( x)(fa Gx) A 2 (2) Fa A (for CP)
45 1 (1) ( x)(fa Gx) A 2 (2) Fa A (for CP) 1 (3) (Fa Gb) 1 E
46 1 (1) ( x)(fa Gx) A 2 (2) Fa A (for CP) 1 (3) (Fa Gb) 1 E 1,2 (4) Gb 2,3 E
47 1 (1) ( x)(fa Gx) A 2 (2) Fa A (for CP) 1 (3) (Fa Gb) 1 E 1,2 (4) Gb 2,3 E 1,2 (5) ( x)gx 4 I
48 1 (1) ( x)(fa Gx) A 2 (2) Fa A (for CP) 1 (3) (Fa Gb) 1 E 1,2 (4) Gb 2,3 E 1,2 (5) ( x)gx 4 I 1 (6) (Fa ( x)gx) 2,5 CP
49 Existential Introduction ( I) Suppose ( x)ϕ is a well-formed formula for some variable x, and suppose c is an arbitrary constant. ϕ[x/c] ( x)ϕ
50 Let s try an example: { Fa } ( x)fx
51 1 (1) Fa A
52 1 (1) Fa A 1 (2) ( x)fx 1 I
53 Let s try an example: { } (Fbb (( x)fxb ᴧ ( x)fbx)
54 1 (1) Fbb A (for CP)
55 1 (1) Fbb A (for CP) 1 (2) ( x)fxb 1 I
56 1 (1) Fbb A (for CP) 1 (2) ( x)fxb 1 I 1 (3) ( x)fbx 1 I
57 1 (1) Fbb A (for CP) 1 (2) ( x)fxb 1 I 1 (3) ( x)fbx 1 I 1 (4) (( x)fxb ᴧ ( x)fbx) 2,3 ᴧI
58 1 (1) Fbb A (for CP) 1 (2) ( x)fxb 1 I 1 (3) ( x)fbx 1 I 1 (4) (( x)fxb ᴧ ( x)fbx) 2,3 ᴧI (5) (1) (4) 1,4 CP
59 And one more: { ( x)( y)rxy } ( x)( y)rxy)
60 1 (1) ( x)( y)rxy A
61 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E
62 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E 1 (3) Rab 2 E
63 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E 1 (3) Rab 2 E 1 (4) ( y)ray 3 I
64 1 (1) ( x)( y)rxy A 1 (2) ( y)ray 1 E 1 (3) Rab 2 E 1 (4) ( y)ray 3 I 1 (5) ( x)( y)rxy 4 I
65 Next Time We ll have a midterm exam. Then next week, we ll look at one more rule for the existential quantifier and two rules for the identity relation.
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