1. (G&H)>(J<>L)
2. G<>H
3. (H&~L)v(H&K)
Conclusion: J>K
----------------------------
Using a conditional proof, we assume the truth of J,
so we put a "t" over all J's
--------------------
t
1. (G&H)>(J<>L)
2. G<>H
3. (H&~L)v(H&K)
t
Conclusion: J>K
---------------------
Since J<>L appears in 1, and J is t, then we must put a t above all Ls
---------------------
t t
1. (G&H)>(J<>L)
2. G<>H
t
3. (H&~L)v(H&K)
t
Conclusion: J>K
---------------------
Since L has a t over it, in the ~L of 3, we put an f over the ~
t t
1. (G&H)>(J<>L)
2. G<>H
ft
3. (H&~L)v(H&K)
t
Conclusion: J>K
---------------------
Since there is an f over the ~L in 3, there is also an f
over the & of (H&~L).
t t
1. (G&H)>(J<>L)
2. G<>H
fft
3. (H&~L)v(H&K)
t
Conclusion: J>K
---------------------
Since the left side of 3. (H&~L) has an f over it, the right
side must have a t over it, so that 3 will be a true premise,
so we put a t over the & of H&K
t t
1. (G&H)>(J<>L)
2. G<>H
fft t t
3. (H&~L)v(H&K)
t
Conclusion: J>K
---------------------
Since the & of H&K has a t over it, both H, K must have t's over them.
So we put t's over all K's
t t
1. (G&H)>(J<>L)
2. G<>H
fft t ttt
3. (H&~L)v(H&K)
t t
Conclusion: J>K
---------------------
Therefore the conclusion is valid, because the assumption of the
truth of J leads to the truth of K.
Edwin
