AND:
OR:
IF…THEN:
IFF:
(P⊃Q) and (Q⊃P) from (P≡Q)
(P v Q) and ~(P·Q) from ~(P≡Q)
Steps:
1. Reverse squiggles (mark the line)
2. Drop existentials (mark the line) - ∃x
3. Drop universals (don't mark the line) – x
∃x become x vice versa:
~(x)Lx → (∃x)~Lx
~(∃x)Lx→(x)~Lx
Add constant:
(∃x)(Rx·Bx) → (Ra·Ba)
However, cannot: (∃x)(Fx·(x)Gx) → (Fa·(x)Gx)
(Ex) Px-> Pa // can only be used once, need to use new constants afterwards
(x)Bx -> Ba // can be used multiple times, use the old constants
1.
Each quantifier
– All are R = (x)Rx
– Not all are R= ~(x)Rx
– Some are R = there is an R = (∃x)Rx,
– Nothing is R = there is no R = ~(∃x)Rx
Derived line generates a contradiction, next line is to infer the negation of the latest
assumption
Test 1:
Test 2:
Test 3:
Test 4:
Test 5:
Test 6:
Test 7:
Test 8: