Last Update: 1 October 2010
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(*) (A ∧ (A → B)) → B)
But it isn't!
is valid iff ((A ∧ B) → C) is a tautology
MP | A → B A B |
¬A ∨ B A B |
||
MT | A → B ¬B ¬A |
¬A ∨ B ¬B ¬A |
||
HS | A → B B → C A → C |
¬A ∨ B ¬B ∨ C ¬A ∨ C | ||
DS | A ∨ B ¬A B |
(already uses only ¬,∨) |
A ∨ B_{1} ∨ … ∨ B_{n}
¬A ∨ C_{1} ∨ … ∨ C_{m}
(B_{1} ∨ … ∨ B_{n})
∨
(C_{1} ∨ … ∨ C_{m})
or, more simply:
A ∨ B
¬A ∨ C
B ∨ C
P1: Lynn works part time or full time.
P2: If Lynn doesn't play on the team,
then she doesn't work part time.
P3: If Lynn plays on the team,
then she's busy.
P4: Lynn doesn't work full time.
C : ∴ Lynn is busy.
Let pt = Lynn works part time.
Let ft = Lynn works full time.
Let plays = Lynn plays on the team.
Let busy = Lynn is busy.
P1: pt &or ft
P2: ¬plays → ¬pt
P3: plays → busy
P4: ¬ft
C : ∴ busy