Solutions to Chapter 38 Conversion of quantifiers

A. Show in each case that the sentences are inconsistent:

  1. 1.
    ​

    S⁢a→T⁡(m),T⁡(m)→S⁡(a),T⁡(m)∧¬S⁡(a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    S⁡(a)→T⁡(m)

    2

    0
    T⁡(m)→S⁡(a)

    3

    0
    T⁡(m)∧¬S⁡(a)

    4

    0
    T⁡(m)
    ∧E 3

    5

    0
    ¬S⁡(a)
    ∧E 3

    6

    0
    S⁡(a)
    →E 2, 4

    7

    0
    ⊥
    ¬E 5, 6
  2. 2.
    ​

    ¬∃x⁢R⁡(x,a),∀x⁢∀y⁢R⁡(y,x)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ¬∃x⁢R⁡(x,a)

    2

    0
    ∀x⁢∀y⁢R⁡(y,x)

    3

    0
    ∀x⁢¬R⁡(x,a)
    CQ 1

    4

    0
    ¬R⁡(b,a)
    ∀E 3

    5

    0
    ∀y⁢R⁡(y,a)
    ∀E 2

    6

    0
    R⁡(b,a)
    ∀E 5

    7

    0
    ⊥
    ¬E 6, 4
  3. 3.
    ​

    ¬∃x⁢∃y⁢L⁡(x,y),L⁡(a,a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ¬∃x⁢∃y⁢L⁡(x,y)

    2

    0
    L⁡(a,a)

    3

    0
    ∀x⁢¬∃y⁢L⁡(x,y)
    CQ 1

    4

    0
    ¬∃y⁢L⁡(a,y)
    ∀E 3

    5

    0
    ∀y⁢¬L⁡(a,y)
    CQ 4

    6

    0
    ¬L⁡(a,a)
    ∀E 5

    7

    0
    ⊥
    ¬E 2, 6
  4. 4.
    ​

    ∀x⁢(P⁡(x)→Q⁡(x)),∀z⁢(P⁡(z)→R⁡(z)),∀y⁢P⁡(y),¬Q⁡(a)∧¬R⁡(b)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(P⁡(x)→Q⁡(x))

    2

    0
    ∀z⁢(P⁡(z)→R⁡(z))

    3

    0
    ∀y⁢P⁡(y)

    4

    0
    ¬Q⁡(a)∧¬R⁡(b)

    5

    0
    ¬Q⁡(a)
    ∧E 4

    6

    0
    P⁡(a)→Q⁡(a)
    ∀E 1

    7

    0
    ¬P⁡(a)
    MT 6, 5

    8

    0
    P⁡(a)
    ∀E 3

    9

    0
    ⊥
    ¬E 8, 7

B. Show that each pair of sentences is interderivable:

  1. 1.
    ​

    ∀x⁢(A⁡(x)→¬B⁡(x)),¬∃x⁢(A⁡(x)∧B⁡(x))

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(A⁡(x)→¬B⁡(x))

    2

    open subproof, 1
    ∃x⁢(A⁡(x)∧B⁡(x))

    3

    open subproof, 2
    A⁡(a)∧B⁡(a)

    4

    2
    A⁡(a)
    ∧E 3

    5

    2
    B⁡(a)
    ∧E 3

    6

    2
    A⁡(a)→¬B⁡(a)
    ∀E 1

    7

    2
    ¬B⁡(a)
    →E 6, 4

    8

    2
    ⊥
    ¬E 5, 7

    9

    close subproof, 1
    ⊥
    ∃E 2, 3–8

    10

    close subproof, 0
    ¬∃x⁢(A⁡(x)∧B⁡(x))
    ¬I 2–9

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ¬∃x⁢(A⁡(x)∧B⁡(x))

    2

    0
    ∀x⁢¬(A⁡(x)∧B⁡(x))
    CQ 1

    3

    0
    ¬(A⁡(a)∧B⁡(a))
    ∀E 2

    4

    open subproof, 1
    A⁡(a)

    5

    open subproof, 2
    B⁡(a)

    6

    2
    A⁡(a)∧B⁡(a)
    ∧I 4, 5

    7

    2
    ⊥
    ¬E 6, 3

    8

    close subproof, 1
    ¬B⁡(a)
    ¬I 5–7

    9

    close subproof, 0
    A⁡(a)→¬B⁡(a)
    →I 4–8

    10

    0
    ∀x⁢(A⁡(x)→¬B⁡(x))
    ∀I 9
  2. 2.
    ​

    ∀x⁢(¬A⁡(x)→B⁡(d)),∀x⁢A⁡(x)∨B⁡(d)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(¬A⁡(x)→B⁡(d))

    2

    0
    ¬A⁡(a)→B⁡(d)
    ∀E 1

    3

    open subproof, 1
    B⁡(d)

    4

    1
    ∀x⁢A⁡(x)∨B⁡(d)
    ∨I 6

    5

    close subproof, open subproof, 1
    ¬B⁡(d)

    6

    1
    ¬¬A⁡(a)
    MT 2, 5

    7

    1
    A⁡(a)
    DNE 6

    8

    1
    ∀x⁢A⁡(x)
    ∀E 7

    9

    1
    ∀x⁢A⁡(x)∨B⁡(d)
    ∨I 8

    10

    close subproof, 0
    ∀x⁢A⁡(x)∨B⁡(d)
    LEM 3–4, 5–9

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢A⁡(x)∨B⁡(d)

    2

    open subproof, 1
    ¬A⁡(a)

    3

    open subproof, 2
    ∀x⁢A⁡(x)

    4

    2
    A⁡(a)
    ∀E 3

    5

    2
    ⊥
    ¬E 4, 2

    6

    close subproof, 1
    ¬∀x⁢A⁡(x)
    ¬I 3–5

    7

    1
    B⁡(d)
    DS 1, 6

    8

    close subproof, 0
    ¬A⁡(a)→B⁡(d)
    →I 2–7

    9

    0
    ∀x⁢(A⁡(x)→B⁡(d))
    ∀I 8

C. In appendix 24, we considered what happens when we move quantifiers ‘across’ various logical operators. Show that each pair of sentences is interderivable:

  1. 1.
    ​

    ∀x⁢(F⁡(x)∧G⁡(a)),∀x⁢F⁡(x)∧G⁡(a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(F⁡(x)∧G⁡(a))

    2

    0
    F⁡(b)∧G⁡(a)
    ∀E 1

    3

    0
    F⁡(b)
    ∧E 2

    4

    0
    G⁡(a)
    ∧E 6

    5

    0
    ∀x⁢F⁡(x)
    ∀I 3

    6

    0
    ∀x⁢F⁡(x)∧G⁡(a)
    ∧I 5, 4

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢F⁡(x)∧G⁡(a)

    2

    0
    ∀x⁢F⁡(x)
    ∧E 1

    3

    0
    G⁡(a)
    ∧E 1

    4

    0
    F⁡(b)
    ∀E 2

    5

    0
    F⁡(b)∧G⁡(a)
    ∧I 4, 3

    6

    0
    ∀x⁢(F⁡(x)∧G⁡(a))
    ∀I 5
  2. 2.
    ​

    ∃x⁢(F⁡(x)∨G⁡(a)),∃x⁢F⁡(x)∨G⁡(a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∃x⁢(F⁡(x)∨G⁡(a))

    2

    open subproof, 1
    F⁡(b)∨G⁡(a)

    3

    open subproof, 2
    F⁡(b)

    4

    2
    ∃x⁢F⁡(x)
    ∃I 3

    5

    2
    ∃x⁢F⁡(x)∨G⁡(a)
    ∨I 4

    6

    close subproof, open subproof, 2
    G⁡(a)

    7

    2
    ∃x⁢F⁡(x)∨G⁡(a)
    ∨I 6

    8

    close subproof, 1
    ∃x⁢F⁡(x)∨G⁡(a)
    ∨E 2, 3–5, 6–7

    9

    close subproof, 0
    ∃x⁢F⁡(x)∨G⁡(a)
    ∃E 1, 2–8

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∃x⁢F⁡(x)∨G⁡(a)

    2

    open subproof, 1
    ∃x⁢F⁡(x)

    3

    open subproof, 2
    F⁡(b)

    4

    2
    F⁡(b)∨G⁡(a)
    ∨I 3

    5

    2
    ∃x⁢(F⁡(x)∨G⁡(a))
    ∃I 4

    6

    close subproof, 1
    ∃x⁢(F⁡(x)∨G⁡(a))
    ∃E 2, 3–5

    7

    close subproof, open subproof, 1
    G⁡(a)

    8

    1
    F⁡(b)∨G⁡(a)
    ∨I 7

    9

    1
    ∃x⁢(F⁡(x)∨G⁡(a))
    ∃I 8

    10

    close subproof, 0
    ∃x⁢(F⁡(x)∨G⁡(a))
    ∨E 1, 2–6, 7–9
  3. 3.
    ​

    ∀x⁢(G⁡(a)→F⁡(x)),G⁡(a)→∀x⁢F⁡(x)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(G⁡(a)→F⁡(x))

    2

    0
    G⁡(a)→F⁡(b)
    ∀E 1

    3

    open subproof, 1
    G⁡(a)

    4

    1
    F⁡(b)
    →E 2, 3

    5

    1
    ∀x⁢F⁡(x)
    ∀I 4

    6

    close subproof, 0
    G⁡(a)→∀x⁢F⁡(x)
    →I 3–5

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    G⁡(a)→∀x⁢F⁡(x)

    2

    open subproof, 1
    G⁡(a)

    3

    1
    ∀x⁢F⁡(x)
    →E 1, 2

    4

    1
    F⁡(b)
    ∀E 3

    5

    close subproof, 0
    G⁡(a)→F⁡(b)
    →I 2–4

    6

    0
    ∀x⁢(G⁡(a)→F⁡(x))
    ∀I 5
  4. 4.
    ​

    ∀x⁢(F⁡(x)→G⁡(a)),∃x⁢F⁡(x)→G⁡(a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢(F⁡(x)→G⁡(a))

    2

    open subproof, 1
    ∃x⁢F⁡(x)

    3

    open subproof, 2
    F⁡(b)

    4

    2
    F⁡(b)→G⁡(a)
    ∀E 1

    5

    2
    G⁡(a)
    →E 4, 3

    6

    close subproof, 1
    G⁡(a)
    ∃E 2, 3–5

    7

    close subproof, 0
    ∃x⁢F⁡(x)→G⁡(a)
    →I 2–6

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∃x⁢F⁡(x)→G⁡(a)

    2

    open subproof, 1
    F⁡(b)

    3

    1
    ∃x⁢F⁡(x)
    ∃I 2

    4

    1
    G⁡(a)
    →E 1, 3

    5

    close subproof, 0
    F⁡(b)→G⁡(a)
    →I 2–4

    6

    0
    ∀x⁢(F⁡(x)→G⁡(a))
    ∀I 5
  5. 5.
    ​

    ∃x⁢(G⁡(a)→F⁡(x)),G⁡(a)→∃x⁢F⁡(x)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∃x⁢(G⁡(a)→F⁡(x))

    2

    open subproof, 1
    G⁡(a)

    3

    open subproof, 2
    G⁡(a)→F⁡(b)

    4

    2
    F⁡(b)
    →E 3, 2

    5

    2
    ∃x⁢F⁡(x)
    ∃I 4

    6

    close subproof, 1
    ∃x⁢F⁡(x)
    ∃E 1, 3–5

    7

    close subproof, 0
    G⁡(a)→∃x⁢F⁡(x)
    →I 2–6

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    G⁡(a)→∃x⁢F⁡(x)

    2

    open subproof, 1
    G⁡(a)

    3

    1
    ∃x⁢F⁡(x)

    4

    open subproof, 2
    F⁡(b)

    5

    open subproof, 3
    G⁡(a)

    6

    3
    F⁡(b)
    R 4

    7

    close subproof, 2
    G⁡(a)→F⁡(b)
    →I 5–6

    8

    2
    ∃x⁢(G⁡(a)→F⁡(x))
    ∃I 7

    9

    close subproof, 1
    ∃x⁢(G⁡(a)→F⁡(x))
    ∃E 3, 4–8

    10

    close subproof, open subproof, 1
    ¬G⁡(a)

    11

    open subproof, 2
    G⁡(a)

    12

    2
    ⊥
    ¬E 11, 10

    13

    2
    F⁡(b)
    X 12

    14

    close subproof, 1
    G⁡(a)→F⁡(b)
    →E 11–13

    15

    1
    ∃x⁢(G⁡(a)→F⁡(x))
    ∃I 14

    16

    close subproof, 0
    ∃x⁢(G⁡(a)→F⁡(x))
    LEM 2–9, 10–15
  6. 6.
    ​

    ∃x⁢(F⁡(x)→G⁡(a)),∀x⁢F⁡(x)→G⁡(a)

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∃x⁢(F⁡(x)→G⁡(a))

    2

    open subproof, 1
    ∀x⁢F⁡(x)

    3

    open subproof, 2
    F⁡(b)→G⁡(a)

    4

    2
    F⁡(b)
    ∀E 2

    5

    2
    G⁡(a)
    →E 3, 4

    6

    close subproof, 1
    G⁡(a)
    ∃E 1, 3–5

    7

    close subproof, 0
    ∀x⁢F⁡(x)→G⁡(a)
    →I 2–6

    Line number

    Subproof level

    Formula

    Justification

    1

    0
    ∀x⁢F⁡(x)→G⁡(a)

    2

    open subproof, 1
    ∀x⁢F⁡(x)

    3

    1
    G⁡(a)
    →E 1, 2

    4

    open subproof, 2
    F⁡(b)

    5

    2
    G⁡(a)
    R 3

    6

    close subproof, 1
    F⁡(b)→G⁡(a)
    →I 4–5

    7

    1
    ∃x⁢(F⁡(x)→G⁡(a))
    ∃I 6

    8

    close subproof, open subproof, 1
    ¬∀x⁢F⁡(x)

    9

    1
    ∃x⁢¬F⁡(x)
    CQ 8

    10

    open subproof, 2
    ¬F⁡(b)

    11

    open subproof, 3
    F⁡(b)

    12

    3
    ⊥
    ¬E 11, 10

    13

    3
    G⁡(a)
    X 12

    14

    close subproof, 2
    F⁡(b)→G⁡(a)
    →I 11–13

    15

    2
    ∃x⁢(F⁡(x)→G⁡(a))
    ∃I 14

    16

    close subproof, 1
    ∃x⁢(F⁡(x)→G⁡(a))
    ∃E 9, 10–15

    17

    close subproof, 0
    ∃x⁢(F⁡(x)→G⁡(a))
    LEM 2–7, 8–16