basic truth tables
how many rows will be in this truth table?
[∼ D & (B ∨ G)] ⊃ [∼ (H & A) ∨ ∼ D]
16
label:
( Fa → ( ∼ Gb → Fa ))
arrow=horseshoe
logically true
is it equivalent? counter example if not
( Gk → Db ) ,( ∼ Gk → ∼ Db )
arrow=horseshoe
TTFT, TFTT,
No, second and third rows
is it valid? show why if not
( Hd ◦ Jc ), ∼ Jc ,( Hd → Jc )
arrow=horseshoe
o=wedge
invalid
TTTF, FTFT, TFTT
second row
does it entail, give a counterexample
( Fa & Gb ), ( ∼ Fa → Gb )
arrow=horseshoe
TFFF, TTTF
Yes
how many rows in this truth table?
(B & C) ⊃ [B ∨ (C & ∼ C)]
4
label:
(( Gb & ∼ Fa ) ◦ Gb )
o=wedge
logically indeterminate
TFTF
is it equivalent?
( Fa & ( Fb ◦ Fc )), (( Fa & Fb ) ◦ ( Fa & Fc ))
o=wedge
YES
TTTFFFFF
is it valid, show why if not
( Fa → Gb ) ,( Gb → Hc ), ( Fa → Hc )
arrow=horseshoe
valid
TTFFTTTT, TFTTFTT, TFTFTTTT
does it entail? give a counter example if not
( Fa → Hc ) ,( Hc → Fa )
arrow=horseshoe
TFTT, TTFT
No, third row
construct a truth table
(A & B) triple bar ∼ B
FFTF
label:
( Gb ↔ ( Fa ◦ ∼( Fa & Gb )))
double arrow=triple bar
circle=wedge
logically indeterminate
is it consistent?
( Fa → Gb ) ,( Fa → ∼ Gb )
show where
arrow=horseshoe
yes
TFTT, FTTT
last two rows
is it valid show why if not
∼( Fa ◦ ( Jd & Fa )), ( Fa ◦ ∼ Jd )
o=wedge
invalid
FFTT, TTFT
3rd row
does it entail:
(( ∼ Hl → Je ) → ( Je → Hl )), ( Hl → Je )
arrow=horseshoe
TTFT, TFTT
No, second row
construct a truth table:
(Ma → ∼ (F g → ∼Eab))
arrow:horseshoe
TTFTFTFT
label:
(( Ha → Jb ) & ∼ ∼(( Jb ◦ Ax ) → Ha ))
arrow: horseshoe
logically contingent
TFFFTFFT
is it consistent?
( Fa → Gb ) ,( Gb → Hc ), ( Fa → ∼ Hc )
show where
arrow-horseshoe
YES
TTFFTTTT, TFTTTFTT, FTFTTTT
first two rows, 6th and last row
is it valid, show why if not
(( Hg ↔ Kd ) ↔ Fa ) ,∼ Hg ,( Kd → ∼ Fa )
double arrow-tripe bar
arrow-horseshoe
valid
TFFTTFFT, FFTTFFTT, FTFTTTTT
Does it entail? give counter example
( Fa & (( Fa ◦ Gb ) → Gb )), Gb
arrow=horseshoe
o=wedge
TFFF, TFTF
Yes
construct a truth table
[Fb → (Gd & Fb)] → ∼J d f )
arrow: horseshoe
FTTTFTFT
LABEL:
Ad Bc Fg (( Ad → ( Bc → Fg )) → (( Ad & Bc ) → Fg ))
arrow: horseshoe
logically true
is it consistent? show where
( Fc → ( Gb → Jm )), ∼ Jm ,( Fc & Jm )
arrow=horseshoe
NO
is it valid show why if not
( Fh → ( ∼ Hf & Kd )) ,( Kd ↔ Hf ), ( Fh → Kd )
VALID
FFTFTTT, TFFTTFFT, TFTFTTTT
Does it entail? give counter example
( Jd ↔ ( Ja → Fc )) ,(( Fc ◦ Ja ) → Ja )
TFFTTFTF, TTTTFFFT,
no, 5th and 7th row