Question 1209282: 1. (~(~Z v H) ⊃ ~T)
2. ((S ⊃ Z) ⊃ ~H)
∴ (Z ⊃ ~T)
what's the next steps for this?
Found 2 solutions by AnlytcPhil, math_tutor2020:
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website!
I am not sure what symbol you use for "and" ("conjunction") so I'll use "&"
1. ~(~Z v H) ⊃ ~T
2. (S ⊃ Z) ⊃ ~H ∴ Z ⊃ ~T
3. ~~T ⊃ ~~(~Z v H) 1, transposition
4. T ⊃ (~Z v H) 4, double negation
5. | 5. ~(Z ⊃ ~T) Assumption for indirect proof
| 6. ~(~~T ⊃ ~Z) 5, Transposition
| 7. ~(T ⊃ ~Z) 6, Double negation
| 8. ~(T ⊃ (~Z v H)) 7, Addition,
| 9. T ⊃ (~Z v H) & ~(T ⊃ (~Z v H))
4,8, Conjunction
10. Z ⊃ ~T lines 5-9 for indirect proof
Notice that premise 2 was not used or needed.
Edwin
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Here is one way to do a derivation using a conditional proof.
I'll use arrow symbols in place of horseshoe symbols.
| Number | Statement | Line(s) Used | Reason | | 1 | | ~(~Z v H) --> ~T | | | | 2 | | (S --> Z) --> ~H | | | | :. | | Z --> ~T | | | | 3 | Z | | Assumption for Conditional Proof | | 4 | (Z & ~H) --> ~T | 1 | De Morgan’s Law | | 5 | Z --> (~H --> ~T) | 4 | Exportation | | 6 | ~H --> ~T | 5,3 | Modus Ponens | | 7 | (S --> Z) --> ~T | 2,6 | Hypothetical Syllogism | | 8 | (~S v Z) --> ~T | 7 | Material Implication | | 9 | Z v ~S | 3 | Addition | | 10 | ~S v Z | 9 | Commutation | | 11 | ~T | 8,10 | Modus Ponens | | 12 | | Z --> ~T | 3 - 11 | Conditional Proof |
I started with assuming Z is the case (line 3). Then I used the logic rules of inference and replacement to arrive at ~T (line 11)
The assumption Z leading to ~T then allows us to prove Z --> ~T is a valid conclusion.
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Here's a way to do it using a direct proof.
| Number | Statement | Line(s) Used | Reason | | 1 | ~(~Z v H) --> ~T | | | | 2 | (S --> Z) --> ~H | | | | :. | Z --> ~T | | | | 3 | (~S v Z) --> ~H | 2 | Material Implication | | 4 | ~(~S v Z) v ~H | 3 | Material Implication | | 5 | (S & ~Z) v ~H | 4 | De Morgan’s Law | | 6 | (S v ~H) & (~Z v ~H) | 5 | Distribution | | 7 | (~Z v ~H) & (S v ~H) | 6 | Commutation | | 8 | ~Z v ~H | 7 | Simplification | | 9 | Z --> ~H | 8 | Material Implication | | 10 | (Z & ~H) --> ~T | 1 | De Morgan’s Law | | 11 | (~H & Z) --> ~T | 10 | Commutation | | 12 | ~H --> (Z --> ~T) | 11 | Exportation | | 13 | Z --> (Z --> ~T) | 9, 12 | Hypothetical Syllogism | | 14 | (Z & Z) --> ~T | 13 | Exportation | | 15 | Z --> ~T | 14 | Tautology |
There might be a much more efficient pathway, but I'm not able to think of it right now.
Caution: Tutor Edwin makes the mistake of using the Addition Rule on part of an expression rather than the entire thing.
It is NOT valid to go from ~(T ⊃ ~Z) to ~(T ⊃ (~Z v H))
Notice in this rule set the "addition" property is in the "rules of inference" sub-block. The rules of inference must apply to the entire line. In contrast a rule of replacement can apply to a portion of a line.
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