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I am totally mixed up with these:

using ONLY this three axioms and Modus Ponens:$$1. \ F \implies (G\implies F) \\ 2. \ (F \implies (G\implies H))\implies ((F \implies G)\implies (F \implies H)) \\ 3. \ (\neg G \implies \neg F)\implies ((\neg G\implies F)\implies G)$$

I need to prove: $$4. \ (\neg G\implies \neg F) \implies (F \implies G)\\ 5. (F \implies G)\implies ((G \implies H)\implies(F\implies H))\\ 6.(F\implies (G \implies H))\implies(G\implies (F\implies H)$$ I see it intuitively, but I have to use ONLY the axioms. I appreciate any help.

Trevor J Richards
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tmac_balla
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2 Answers2

2

In oder to prove 5 and 6, we need some preliminary results.

T1 : $P \rightarrow P$

1) $P \rightarrow ((Q \rightarrow P) \rightarrow P)$ --- Ax.1

2) $P \rightarrow (Q \rightarrow P)$ --- Ax.1

3) $(1) \rightarrow ((2) \rightarrow (P \rightarrow P))$ --- Ax.2

4) $P \rightarrow P$ --- from 3), 1) and 2) by Modus Ponens twice.

T2 : $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$

1) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2

2) $(1) \rightarrow ((Q \rightarrow R) \rightarrow (1))$ --- Ax.1

3) $(Q \rightarrow R) \rightarrow (1)$ --- from 1) and 2) by Modus Ponens

4) $(Q \rightarrow R) \rightarrow (P \rightarrow (Q \rightarrow R))$ --- Ax.1

5) $(3) \rightarrow ((4) \rightarrow ((Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))))$ --- Ax.2

6) $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- from 5), 3) and 4) by MP twice.

T3 : $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$

1) $(P \rightarrow Q) \rightarrow (P \rightarrow Q)$ --- T1

2) $(1) \rightarrow (((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q))$ --- Ax.2

3) $((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 1) and 2) by MP

4) $P \rightarrow ((P \rightarrow Q) \rightarrow P)$ --- Ax.1

5) $(3) \rightarrow ((4) \rightarrow (P \rightarrow ((P \rightarrow Q) \rightarrow Q)))$ --- T2

6) $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 5), 3) and 4) by MP twice.

T4 : $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$

1) $((P \rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow ((Q \rightarrow (P \rightarrow Q)) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- T2

2) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2

3) $Q \rightarrow (P \rightarrow Q)$ --- Ax.1

4) $(1) \rightarrow ((2) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))))$ --- T2

5) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- from 4), 1) and 2) by MP twice

6) $(3) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- T3

7) $((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 6) and 3) by MP

8) $(7) \rightarrow ((5) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))))$ --- T2

9) $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 8), 7) and 5) by MP twice.

This is 6 above.

T5 : $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$

1) $((Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))) \rightarrow ((P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R)))$ --- T4

2) $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$ --- from 1) and T2 by MP.

This is 5 above.

Mauro ALLEGRANZA
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1

To prove 4...

Step 1.) Suppose $\neg G\Longrightarrow\neg F$. (We wish to finally conclude $F\Longrightarrow G$.)

Step 2.) Then $((\neg G\Longrightarrow F)\Longrightarrow G$ (by $(3)$).

Step 3.) Suppose now $F$.

Step 4.) Then $\neg G\Longrightarrow F$ (by Step $3$ and $(1)$).

Step 5.) Then $G$ (by Step (4), Step $(2)$, and Modes Ponens).

Step 6.) Then $F\Longrightarrow G$ (by Step $3$ and Step $5$).

Conclude finally that $(\neg G\Longrightarrow\neg F)\Longrightarrow(F\Longrightarrow G)$ (by Step $1$ and Step $6$).

Trevor J Richards
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