Solutions to Chapter 17 Basic rules for TFL

A. The following two ‘proofs’ are incorrect. Explain the mistakes they make.

$\wedge$E on line 4 can’t be applied to line 1, since it is not of the form $\mathrm{𝒜︁}\wedge \mathrm{ℬ︁}$. ‘$A$’ could be obtained by $\wedge$E, but from line 2.

$\wedge$E on line 3 should yield ‘$B\wedge C$’. ‘$B$’ could then be obtained by $\wedge$E again.

B. The following three proofs are missing their citations (rule and line numbers). Add them, to turn them into bona fide proofs. Additionally, write down the argument that corresponds to each proof.

Corresponding argument:
$P\wedge S,S\to R\therefore R\vee E$

Corresponding argument:
$A\to D\therefore (A\wedge B)\to (D\vee E)$

Corresponding argument:
$\neg L\to (J\vee L),\neg L\therefore J$

C. Give a proof for each of the following arguments:

1. 1.
   ​

   $J\to \neg J\therefore \neg J$

   Line number	Subproof level	Formula	Justification
   $1$	0	$J\to \neg J$
   $2$	open subproof, 1	$J$
   $3$	1	$\neg J$	$\to$E $1$, $2$
   $4$	1	$\perp$	$\neg$E $2$, $3$
   $5$	close subproof, 0	$\neg J$	$\neg$I $2$–$4$

2. 2.
   ​

   $Q\to (Q\wedge \neg Q)\therefore \neg Q$

   Line number	Subproof level	Formula	Justification
   $1$	0	$Q\to (Q\wedge \neg Q)$
   $2$	open subproof, 1	$Q$
   $3$	1	$Q\wedge \neg Q$	$\to$E $1$, $2$
   $4$	1	$\neg Q$	$\wedge$E $3$
   $5$	1	$\perp$	$\neg$E $2$, $4$
   $6$	close subproof, 0	$\neg Q$	$\neg$I $2$–$5$

3. 3.
   ​

   $A\to (B\to C)\therefore (A\wedge B)\to C$

   Line number	Subproof level	Formula	Justification
   $1$	0	$A\to (B\to C)$
   $2$	open subproof, 1	$A\wedge B$
   $3$	1	$A$	$\wedge$E $2$
   $4$	1	$B\to C$	$\to$E $1$, $3$
   $5$	1	$B$	$\wedge$E $2$
   $6$	1	$C$	$\to$E $4$, $5$
   $7$	close subproof, 0	$(A\wedge B)\to C$	$\to$I $2$–$6$

4. 4.
   ​

   $K\wedge L\therefore K\leftrightarrow L$

   Line number	Subproof level	Formula	Justification
   $1$	0	$K\wedge L$
   $2$	open subproof, 1	$K$
   $3$	1	$L$	$\wedge$E $1$
   $4$	close subproof, open subproof, 1	$L$
   $5$	1	$K$	$\wedge$E $1$
   $6$	close subproof, 0	$K\leftrightarrow L$	$\leftrightarrow$I $2$–$3$, $4$–$5$

5. 5.
   ​

   $(C\wedge D)\vee E\therefore E\vee D$

   Line number	Subproof level	Formula	Justification
   $1$	0	$(C\wedge D)\vee E$
   $2$	open subproof, 1	$C\wedge D$
   $3$	1	$D$	$\wedge$E $2$
   $4$	1	$E\vee D$	$\vee$I $3$
   $5$	close subproof, open subproof, 1	$E$
   $6$	1	$E\vee D$	$\vee$I $5$
   $7$	close subproof, 0	$E\vee D$	$\vee$E $1$, $2$–$4$, $5$–$6$

6. 6.
   ​

   $A\leftrightarrow B,B\leftrightarrow C\therefore A\leftrightarrow C$

   Line number	Subproof level	Formula	Justification
   $1$	0	$A\leftrightarrow B$
   $2$	0	$B\leftrightarrow C$
   $3$	open subproof, 1	$A$
   $4$	1	$B$	$\leftrightarrow$E $1$, $3$
   $5$	1	$C$	$\leftrightarrow$E $2$, $4$
   $6$	close subproof, open subproof, 1	$C$
   $7$	1	$B$	$\leftrightarrow$E $2$, $6$
   $8$	1	$A$	$\leftrightarrow$E $1$, $7$
   $9$	close subproof, 0	$A\leftrightarrow C$	$\leftrightarrow$I $3$–$5$, $6$–$8$

7. 7.
   ​

   $\neg F\to G,F\to H\therefore G\vee H$

   Line number	Subproof level	Formula	Justification
   $1$	0	$\neg F\to G$
   $2$	0	$F\to H$
   $3$	open subproof, 1	$\neg (G\vee H)$
   $4$	open subproof, 2	$F$
   $5$	2	$H$	$\to$E $2$, $4$
   $6$	2	$G\vee H$	$\vee$I $5$
   $7$	2	$\perp$	$\neg$E $3$, $6$
   $8$	close subproof, 1	$\neg F$	$\neg$I $4$–$7$
   $9$	1	$G$	$\to$E $1$, $8$
   $10$	1	$G\vee H$	$\vee$I $9$
   $11$	1	$\perp$	$\neg$E $3$, $10$
   $12$	close subproof, 0	$G\vee H$	IP $3$–$11$

8. 8.
   ​

   $(Z\wedge K)\vee (K\wedge M),K\to D\therefore D$

   Line number	Subproof level	Formula	Justification
   $1$	0	$(Z\wedge K)\vee (K\wedge M)$
   $2$	0	$K\to D$
   $3$	open subproof, 1	$Z\wedge K$
   $4$	1	$K$	$\wedge$E $3$
   $5$	close subproof, open subproof, 1	$K\wedge M$
   $6$	1	$K$	$\wedge$E $5$
   $7$	close subproof, 0	$K$	$\vee$E $1$, $3$–$4$, $5$–$6$
   $8$	0	$D$	$\to$E $2$, $7$

9. 9.
   ​

   $P\wedge (Q\vee R),P\to \neg R\therefore Q\vee E$

   Line number	Subproof level	Formula	Justification
   $1$	0	$P\wedge (Q\vee R)$
   $2$	0	$P\to \neg R$
   $3$	0	$P$	$\wedge$E $1$
   $4$	0	$\neg R$	$\to$E $2$, $3$
   $5$	0	$Q\vee R$	$\wedge$E $1$
   $6$	open subproof, 1	$Q$
   $7$	1	$Q\vee E$	$\vee$I $6$
   $8$	close subproof, open subproof, 1	$R$
   $9$	1	$\perp$	$\neg$E $8$, $4$
   $10$	1	$Q\vee E$	X $9$
   $11$	close subproof, 0	$Q\vee E$	$\vee$E $5$, $6$–$7$, $8$–$10$

10. 10.
    ​

    $S\leftrightarrow T\therefore S\leftrightarrow (T\vee S)$

    Line number	Subproof level	Formula	Justification
    $1$	0	$S\leftrightarrow T$
    $2$	open subproof, 1	$S$
    $3$	1	$T$	$\leftrightarrow$E $1$, $2$
    $4$	1	$T\vee S$	$\vee$I $3$
    $5$	close subproof, open subproof, 1	$T\vee S$
    $6$	open subproof, 2	$T$
    $7$	2	$S$	$\leftrightarrow$E $1$, $6$
    $8$	close subproof, open subproof, 2	$S$
    $9$	2	$S\wedge S$	$\wedge$I $8$, $8$
    $10$	2	$S$	$\wedge$E $9$
    $11$	close subproof, 1	$S$	$\vee$E $5$, $6$–$7$, $8$–$10$
    $12$	close subproof, 0	$S\leftrightarrow (T\vee S)$	$\leftrightarrow$I $2$–$4$, $5$–$11$

11. 11.
    ​

    $\neg (P\to Q)\therefore \neg Q$

    Line number	Subproof level	Formula	Justification
    $1$	0	$\neg (P\to Q)$
    $2$	open subproof, 1	$Q$
    $3$	open subproof, 2	$P$
    $4$	2	$Q\wedge Q$	$\wedge$I $2$, $2$
    $5$	2	$Q$	$\wedge$E $4$
    $6$	close subproof, 1	$P\to Q$	$\to$I $3$–$5$
    $7$	1	$\perp$	$\neg$E $6$, $1$
    $8$	close subproof, 0	$\neg Q$	$\neg$I $2$–$7$

12. 12.
    ​

    $\neg (P\to Q)\therefore P$

    Line number	Subproof level	Formula	Justification
    $1$	0	$\neg (P\to Q)$
    $2$	open subproof, 1	$\neg P$
    $3$	open subproof, 2	$P$
    $4$	2	$\perp$	$\neg$E $3$, $2$
    $5$	2	$Q$	X $4$
    $6$	close subproof, 1	$P\to Q$	$\to$I $3$–$5$
    $7$	1	$\perp$	$\neg$E $6$, $1$
    $8$	close subproof, 0	$P$	IP $2$–$7$