Question 1

select the best English interpretation of the given statements of predicate logic, using the following translation key:
t: two
Ox: x is odd 
Ex: x is even
Nx: x is a number 
Gxy: x is greater than y
-(∀x)[Nx ⊃ (Ox ⊃ ∼Ex)]
A) All numbers are odd if they are not even.
B) All numbers are odd or even.
C) No even number is odd.
D) No odd number is even.
E) If a number is odd it is even.

Accepted Answer

The given statement translates to "For all x, if x is a number, then if x is odd, x is not even," which directly means that no odd number is even.

Question 2

select the best English interpretation of the given statements of predicate logic, using the following translation key:
t: two
Ox: x is odd 
Ex: x is even
Nx: x is a number 
Gxy: x is greater than y
-∼(∃x)(Nx • Gxx)
A) Some numbers are not greater than themselves.
B) Some numbers are not greater than any other number.
C) No number is greater than itself.
D) No number is greater than any other number.
E) Some number is no greater than itself.

Accepted Answer

The statement can be rewritten as "For all x, if x is a number, then x is not greater than itself." This is the definition of reflexivity in mathematics, which states that every object is equal to itself. Therefore, the correct interpretation is that no number is greater than itself.

Question 3

select the best English interpretation of the given statements of predicate logic, using the following translation key:
t: two
Ox: x is odd 
Ex: x is even
Nx: x is a number 
Gxy: x is greater than y
-~(∃x)[(Nx • ∼Ex) • Gxt]
A) No number that is not even is greater than two.
B) Some even numbers are not greater than two.
C) Some number greater than two is not even.
D) Two is the smallest even number.
E) Two is not the smallest even number.

Accepted Answer

The given statement translates to "There does not exist a number x such that x is a number, x is not even, and x is greater than two." This means no number that is odd (since it's not even) and is a number, is greater than two, which matches choice A.

Question 4

select the best English interpretation of the given statements of predicate logic, using the following translation key:
t: two
Ox: x is odd 
Ex: x is even
Nx: x is a number 
Gxy: x is greater than y
-(∀x){(Nx • Ox) ⊃ (∃y)[(Ey • Ny) • Gyx)]}
A) For every odd number, there is an even number greater than it.
B) Every odd number is greater than some even number.
C) All odd numbers are smaller than all even numbers.
D) Any number smaller than every odd number is even.
E) Any number greater than every odd number is even.

Accepted Answer

The statement is a universal quantification (∀x) of the conditional statement [(Nx • Ox) ⊃ (∃y)[(Ey • Ny) • Gyx)], which can be translated as "If x is an odd number, then there exists an even number y such that y is greater than x." Therefore, option A is the best interpretation of the given statement.

Question 5

select the best English interpretation of the given statements of predicate logic, using the following translation key:
t: two
Ox: x is odd 
Ex: x is even
Nx: x is a number 
Gxy: x is greater than y
-(∀x){(Ex • Nx) ⊃ (∀y)[(Oy • Ny) ⊃ ∼Gxy]}
A) All even numbers are smaller than all odd numbers.
B) No even number is greater than any odd number.
C) No odd number is greater than any even number.
D) All even numbers are greater than all odd numbers.
E) No even number is greater than every odd number.

Accepted Answer

The answer of select the best English interpretation of the...

Question 6

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} a = Mercury b = Jupiter c = Saturn d = Pluto Mx = {Mercury, Mars} Px = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune} Cxy = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) Pa • Pd Cad • ~Cda B) Pb • Pd Cbd • ~Cdb C) Pa • Pb Cab • ∼Cba D) Pb • Pa Cba • ~Cab E) Pc • Pb Ccb • ~Cbc

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 7

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} a = Mercury b = Jupiter c = Saturn d = Pluto Mx = {Mercury, Mars} Px = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune} Cxy = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (Ma • Pb) • (Pc $ \lor $ Pd) (∃x)Cxa • ~(∃x)Cxd B) (Pa • Mb) • (Pc $ \lor $ Pd) (∃x)Cxd $ \lor $ ~(∃x)Cxa C) (Pa • Pb) • (Pc $ \lor $ Pd) (∃x)Cxd • ∼(∃x)Cxa D) (Pa • Pb) • (Mc $ \lor $ Pd) (∃x)Cxd $ \lor $ ~(∃x)Cxa E) (Pa • Pb) • (Pc $ \lor $ Md) (∃x)Cxa • ~(∃x)Cxd

Accepted Answer

We need to evaluate the given assignments set to the predicates and check which theory it models. 
Option A contains the clauses (∃x)Cxa and ~(∃x)Cxd, which are not true in the assignment. 
Option B contains the clause (∃x)Cxd, which is not true in the assignment. 
Option D contains the clause (∃x)Cxd, which is not true in the assignment. 
Option E contains the clauses (∃x)Cxa and ~(∃x)Cxd, which are not true in the assignment. 
Thus, option C remains. The interpretation satisfies (Pa • Pb) • (Pc $ \lor $ Pd) which is true in the given assignment. It also satisfies (∃x)Cxd • ∼(∃x)Cxa, which is also true in the given assignment. Therefore, the correct answer is C.

Question 8

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} a = Mercury b = Jupiter c = Saturn d = Pluto Mx = {Mercury, Mars} Px = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune} Cxy = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) Ma • ~Mc (∃x)Cxc ∼(∃x)(∼Cxc • Mx) B) Ma • Mc (∃x)Ccx (∃x)(~Cxc • Mx) C) ~(Ma • Mc) (∃x)Ccx ~(∃x)(Cxc • Mx) D) ~(Ma $ \lor $ Mc) (∃x)Ccx ~(∃x)(Ccx • Mx) E) Ma ≡ ~Mc (∃x)Cax ~(∃x)(Cxc • ~Mx)

Accepted Answer

The interpretation matches theory A because: - Ma is true (Mercury is in Mx), Mc is false (Saturn is not in Mx), there exists an x such that Cxc is true (many pairs in Cxy have Saturn as the second element), and there does not exist an x such that ~Cxc and Mx are both true (all elements related to Saturn in Cxy are not in Mx, and all elements in Mx are related to Saturn in Cxy).

Question 9

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} a = Mercury b = Jupiter c = Saturn d = Pluto Mx = {Mercury, Mars} Px = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune} Cxy = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (∀x){Px ⊃ [(∃y)Cxy • (∃y)Cyx]} Cab • ~Cba Cca • ~Cac Cad • ~Cda B) (∀x){Px ⊃ [(∃y)Cxy ≡ (∃x)Cyx]} Cab • ~Cba Cac • ~Cca Cad • ~Cda C) (∀x)[Px ⊃ (∃y)(Cxy • Cyx)] Cba • ~Caa Cca • ~Ccc Cda • ~Cdd D) (∀x){Px ⊃ [(∃y)Cxy $ \lor $ (∃y)Cyx]} Cab • ∼Cba Cac • ∼Cca Cad • ∼Cda E) (∀x)[Px ⊃ (∃y)(Cxy $ \lor $ Cyx)] Cab • ~Cba Cac • Cad ~(~Cca • ~Cda)

Accepted Answer

The interpretation provided models theory D correctly. The universal quantifier in D states that for all x in Px, there exists some y such that Cxy or Cyx holds, which aligns with the given assignments. The specific instances of Cab, ~Cba, Cac, ~Cca, and Cad, ~Cda are all supported by the assignments to Cxy, showing that for each pair, one of the relations holds but not the inverse, which matches the logical disjunction in the theory.

Question 10

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} a = Mercury b = Jupiter c = Saturn d = Pluto Mx = {Mercury, Mars} Px = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune} Cxy = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (∀x)(Px ⊃ Mx) ~(∃x)Px (∀x)(∀y)[(Px • Py) ⊃ Cxy] B) (∀x)(Mx ≡ Px) ~(∀x)Mx (∀x)[(∀y)Cxy ⊃ Py] C) (∀x)(Mx $ \lor $ Px) ~(∃x)Mx (∀x)[Px ⊃ ~(∀y)(Py ⊃ Cxy)] D) (∀x)(~Mx $ \lor $ Px) ~(∀x)(Px (∀x)[Px ⊃ (∀y)(Py ⊃ Cxy)] E) (∀x)(Mx ⊃ Px) ∼(∀x)Px (∀x)[∼Px ⊃ (∀y)(Py ⊃ Cyx)]

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 11

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {1, 2, 3, ..., 28, 29, 30} a = 1 e = 21 b = 2 f = 23 c = 4 g = 27 d = 20 h = 29 Ex = {2, 4, 6, ..., 28, 30} Ox = {1, 3, 5, ..., 27, 29} Px = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Sxyz = The set of all triples such that the first is the sum of the second and third {<2, 1, 1>, <3, 1, 2>, <3, 2, 1>, <4, 1, 3>, <4, 2, 2>, <4, 3, 1>, <5, 1, 4>, ... } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) Sdca • Shgb (∃x)(∃y)Sdxy • ~(∃x)(∃y)Sgxy B) Scbb • Seda (∃x)(∃y)Scxy • ~(∃x)(∃y)Sbxy C) Secb • Shga (∃x)(∃y)Sgxy • ~(∀x)(∀y)Sfxy D) Sfeb • Sgfc (∃x)(∃y)Shxy • ∼(∀x)(∀y)Shxy E) Shgb • Sfbe (∃x)(∃y)Sgxy • ~(∃x)Sxcd

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 12

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {1, 2, 3, ..., 28, 29, 30} a = 1 e = 21 b = 2 f = 23 c = 4 g = 27 d = 20 h = 29 Ex = {2, 4, 6, ..., 28, 30} Ox = {1, 3, 5, ..., 27, 29} Px = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Sxyz = The set of all triples such that the first is the sum of the second and third {<2, 1, 1>, <3, 1, 2>, <3, 2, 1>, <4, 1, 3>, <4, 2, 2>, <4, 3, 1>, <5, 1, 4>, ... } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (∃x)(Ex • Sfex) (∀x)(Ox ⊃ ∼Sfex) (∃x)(∃y)(Oy • Sxey) B) (∃x)(Ex • Sefx) (∀x)(Ox ⊃ ~Sefx) (∃x)(∃y)(Oy • Sxfy) C) (∃x)(Ox • Sfex) (∀x)(Ex ⊃ ~Sfex) (∃x)(∃y)(Ey • Sxey) D) (∃x)(Ox • Sefx) (∀x)(Ex ⊃ ~Sefx) (∃x)(∃y)(Ey • Sxfy) E) (∃x)(Ox • Sfxe) (∀x)(Ex ⊃ ~Sfxe) (∃x)(∃y)(Ey • Sfxy)

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 13

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {1, 2, 3, ..., 28, 29, 30} a = 1 e = 21 b = 2 f = 23 c = 4 g = 27 d = 20 h = 29 Ex = {2, 4, 6, ..., 28, 30} Ox = {1, 3, 5, ..., 27, 29} Px = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Sxyz = The set of all triples such that the first is the sum of the second and third {<2, 1, 1>, <3, 1, 2>, <3, 2, 1>, <4, 1, 3>, <4, 2, 2>, <4, 3, 1>, <5, 1, 4>, ... } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) Sgfc • Sgcf (∀x)(∀y)(∀z)(Sxyz ≡ Sxzy) B) Sgba • Sgab (∀x)(∀y)[(∀z)Sxyz ≡ (∀z)Sxzy] C) Seda • Sead (∀x)(∀y)(∀z)(Sxyz ≡ Szyx) D) Sfea • Sfae (∀x)(∀y)(∀z)(Sxyz ≡ Syxz) E) Sdba • Sdab (∀x)(∀y)[(∀z)(Sxyz ≡ (∀z)Syxz]

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 14

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {1, 2, 3, ..., 28, 29, 30} a = 1 e = 21 b = 2 f = 23 c = 4 g = 27 d = 20 h = 29 Ex = {2, 4, 6, ..., 28, 30} Ox = {1, 3, 5, ..., 27, 29} Px = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Sxyz = The set of all triples such that the first is the sum of the second and third {<2, 1, 1>, <3, 1, 2>, <3, 2, 1>, <4, 1, 3>, <4, 2, 2>, <4, 3, 1>, <5, 1, 4>, ... } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (∃x)(∃y)[(Ex • Ey) • Sfxy] (∃x)(∃y)[(Ox • Oy) • Sfxy] (∀x){Ex ⊃ {(∃y)(∃z)[(Ey • Ez) • Sxyz)] $ \lor $ (∃y)(∃z)[(Ox • Oy) • Sxyz]}} B) (∃x)(∃y)[(Ex • Ey) • Sdxy] (∃x)(∃y)[(Ox • Oy) • Sdxy] (∀x){Ex ⊃ {(∃y)(∃z)[(Ey • Ez) • Sxyz)] $ \lor $ (∃y)(∃z)[(Ox • Oy) • Sxyz]}} C) (∃x)(∃y)[(Ex • Ey) • Sdxy] (∃x)(∃y)[(Ox • Oy) • Sdxy] (∀x){Ex ⊃ {(∃y)(∃z)[(Ey • Ez) • Sxyz)] • (∃y)(∃z)[(Ox • Oy) • Sxyz]}} D) (∃x)(∃y)[(Ex • Ey) • Sfxy] (∃x)(∃y)[(Ox • Oy) • Sfxy] (∀x){Ex ⊃ {(∃y)(∃z)[(Ey • Ez) • Sxyz)] • (∃y)(∃z)[(Ox • Oy) • Sxyz]}} E) (∃x)(∃y)[(Ex • Ey) • Sdxy] (∃x)(∃y)[(Ox • Oy) • Sdxy] (∀x){Ox ⊃ {(∃y)(∃z)[(Ey • Ez) • Sxyz)] $ \lor $ (∃y)(∃z)[(Ox • Oy) • Sxyz]}}

Accepted Answer

Theory B is modeled by the given interpretation because it correctly applies the definitions and assignments provided. The first two existential statements in B can be satisfied by choosing appropriate even and odd numbers from the domain that sum to d (20), and the universal statement correctly represents the property that for any even number, there exist pairs of even or odd numbers whose sum matches any given even number in the domain.

Question 15

consider the following domain, assignment of objects in the domain, and assignments sets to predicates. Domain = {1, 2, 3, ..., 28, 29, 30} a = 1 e = 21 b = 2 f = 23 c = 4 g = 27 d = 20 h = 29 Ex = {2, 4, 6, ..., 28, 30} Ox = {1, 3, 5, ..., 27, 29} Px = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Sxyz = The set of all triples such that the first is the sum of the second and third {<2, 1, 1>, <3, 1, 2>, <3, 2, 1>, <4, 1, 3>, <4, 2, 2>, <4, 3, 1>, <5, 1, 4>, ... } -Given the customary truth tables, which of the following theories is modeled by the above interpretation? A) (∃x)(∀y)(∀z)Sxyz (∃x)(∀y)(∀z)Syxz B) (∃x)(∀y)(∀z)∼Sxyz (∃x)(∀y)(∀z)Syxz C) (∃x)(∀y)(∀z)Sxyz (∃x)(∀y)(∀z)∼Syxz D) (∃x)(∀y)∼Sxyy (∃x)(∀y)Syxy E) (∃x)(∀y)(∀z)∼Sxyz (∃x)(∀y)(∀z)∼Syxz

Accepted Answer

The answer of consider the following domain, assignment of objects...

Question 16

provide a conterexample in a finite domain to each given invalid argument.
-1. Ad ⊃ (∀x)Fdx / (∃x)Fdx
A) Counterexample in a domain of one member, in which: Ad: True Fdd: False
B) Counterexample in a domain of one member, in which: Ad: False Fdd: False
C) Counterexample in a domain of one member, in which: Ad: True Fdd: True
D) Counterexample in a domain of two members, in which: Ad: True Fdd: False
Ab: True Fdb: False
E) Counterexample in a domain of two members, in which: Ad: False Fdd: True
Ab: False Fdb: False

Accepted Answer

The answer of provide a conterexample in a finite domain...

Question 17

provide a conterexample in a finite domain to each given invalid argument.
-1. (∃x)(Dxa • Ex)
2) (∃x)(Dxa • Fx) / (∃x)(Ex • Fx)
A) Counterexample in a domain of one member, in which: Ea: True Fa: False Daa: True
B) Counterexample in a domain of one member, in which: Ea: False Fa: True Daa: False
C) Counterexample in a domain of two members, in which: Ea: True Fa: True Daa: True
Eb: True Fb: False Dba: False
D) Counterexample in a domain of two members, in which: Ea: True Fa: False Daa: False
Eb: False Fb: False Dba: True
E) Counterexample in a domain of two members, in which: Ea: True Fa: False Daa: True
Eb: False Fb: True Dba: True

Accepted Answer

The answer of provide a conterexample in a finite domain...

Question 18

provide a conterexample in a finite domain to each given invalid argument.
-1. (∀x)[Hx ⊃ (∃y)(Ix • Jxy)]
2) Ha
3) Ib / Jab
A) Counterexample in a domain of two members, in which:
Ha: True Ia: True Jaa: True Jba: False
Hb: True Ib: False Jab: False Jbb: True
B) Counterexample in a domain of two members, in which:
Ha: True Ia: False Jaa: True Jba: False
Hb: False Ib: True Jab: False Jbb: True
C) Counterexample in a domain of two members, in which:
Ha: True Ia: True Jaa: True Jba: True
Hb: False Ib: True Jab: False Jbb: True
D) Counterexample in a domain of two members, in which:
Ha: False Ia: True Jaa: False Jba: True
Hb: True Ib: True Jab: True Jbb: True
E) Counterexample in a domain of two members, in which:
Ha: True Ia: True Jaa: False Jba: False
Hb: True Ib: False Jab: True Jbb: True

Accepted Answer

The answer of provide a conterexample in a finite domain...

Question 19

provide a conterexample in a finite domain to each given invalid argument.
-1. (∀x)[Px ⊃ (∃y)Qxy]
2) (∀x)[(∃y)Qxy ⊃ (∃y)Rxy]
3) Pa / Rab
A) Counterexample in a domain of two members, in which:
Pa: True Qaa: False Qba: True Raa: True Rba: False
Pb: False Qab: False Qbb: True Rab: False Rbb: True
B) Counterexample in a domain of two members, in which:
Pa: True Qaa: True Qba: False Raa: True Rba: True
Pb: False Qab: True Qbb: False Rab: False Rbb: False
C) Counterexample in a domain of two members, in which:
Pa: True Qaa: False Qba: False Raa: False Rba: True
Pb: True Qab: True Qbb: False Rab: False Rbb: True
D) Counterexample in a domain of two members, in which:
Pa: True Qaa: True Qba: True Raa: True Rba: False
Pb: False Qab: False Qbb: True Rab: True Rbb: True
E) Counterexample in a domain of two members, in which:
Pa: True Qaa: True Qba: True Raa: False Rba: False
Pb: True Qab: False Qbb: False Rab: False Rbb: False

Accepted Answer

The answer of provide a conterexample in a finite domain...

Question 20

provide a conterexample in a finite domain to each given invalid argument.
-1. (∀x)[(∃y)(Fxy • Gy) ⊃ (∀y)(Gy ⊃ Fxy)]
2) Ga / (∀x)(Fxa ⊃ Fax)
A) Counterexample in a domain of two members, in which: Ga: True Faa: True Fba: True
Gb: False Fab: False Fbb: True
B) Counterexample in a domain of two members, in which: Ga: True Faa: False Fba: True
Gb: True Fab: False Fbb: False
C) Counterexample in a domain of two members, in which: Ga: True Faa: True Fba: True
Gb: False Fab: True Fbb: False
D) Counterexample in a domain of two members, in which: Ga: False Faa: True Fba: False
Gb: False Fab: True Fbb: True
E) Counterexample in a domain of two members, in which: Ga: True Faa: False Fba: True
Gb: True Fab: True Fbb: False

Accepted Answer

The counterexample in option A shows that the argument is invalid because the premise (∃y)(Fxy • Gy) is true for x = a and y = b, but the conclusion (∀y)(Gy ⊃ Fxy) is false because Gy is false for y = b and Fby is false.

Quiz 5: Full First-Order Logic

select the best English interpretation of the given statements of predicate logic, using the following translation key: t: two Ox: x is odd Ex: x is even Nx: x is a number Gxy: x is greater than y -(∀x) [Nx ⊃ (Ox ⊃ ∼Ex) ]

select the best English interpretation of the given statements of predicate logic, using the following translation key: t: two Ox: x is odd Ex: x is even Nx: x is a number Gxy: x is greater than y -∼(∃x) (Nx • Gxx)

select the best English interpretation of the given statements of predicate logic, using the following translation key: t: two Ox: x is odd Ex: x is even Nx: x is a number Gxy: x is greater than y -~(∃x) [(Nx • ∼Ex) • Gxt]

select the best English interpretation of the given statements of predicate logic, using the following translation key: t: two Ox: x is odd Ex: x is even Nx: x is a number Gxy: x is greater than y -(∀x) {(Nx • Ox) ⊃ (∃y) [(Ey • Ny) • Gyx) ]}

select the best English interpretation of the given statements of predicate logic, using the following translation key: t: two Ox: x is odd Ex: x is even Nx: x is a number Gxy: x is greater than y -(∀x) {(Ex • Nx) ⊃ (∀y) [(Oy • Ny) ⊃ ∼Gxy]}

provide a conterexample in a finite domain to each given invalid argument. -1. Ad ⊃ (∀x) Fdx / (∃x) Fdx

provide a conterexample in a finite domain to each given invalid argument. -1. (∃x) (Dxa • Ex) 2) (∃x) (Dxa • Fx) / (∃x) (Ex • Fx)

provide a conterexample in a finite domain to each given invalid argument. -1. (∀x) [Hx ⊃ (∃y) (Ix • Jxy) ] 2) Ha 3) Ib / Jab

provide a conterexample in a finite domain to each given invalid argument. -1. (∀x) [Px ⊃ (∃y) Qxy] 2) (∀x) [(∃y) Qxy ⊃ (∃y) Rxy] 3) Pa / Rab

provide a conterexample in a finite domain to each given invalid argument. -1. (∀x) [(∃y) (Fxy • Gy) ⊃ (∀y) (Gy ⊃ Fxy) ] 2) Ga / (∀x) (Fxa ⊃ Fax)