Proving $\forall x \forall y(R(x,y) \to R(y,x)) \therefore \forall x \forall y\forall z[(R(x,y) \wedge R(x,z)) \to \exists u(R(y,u) \wedge R(z,u))]$

Question

Working on Working on P.D. Magnus. forallX: an Introduction to Formal Logic (pp. 297, exercise B. 4), appears this exercise:

$ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{\forall x \forall y(R(x,y) \to R(y,x))}{ \fitch{R(a,b) \wedge R(a,c)}{ R(a,b) \to R(b,a)\\ R(a,c) \to R(c,a)\\ R(a,b)\\ R(b,a)\\ R(a,c)\\ R(c,a)\\ R(b,a) \wedge R(c,a)\\ \exists u(R(b,u) \wedge R(c,u)) }\\ (R(a,b) \wedge R(a,c)) \to \exists u(R(b,u) \wedge R(c,u))\\ \forall z[(R(a,b) \wedge R(a,z)) \to \exists u(R(b,u) \wedge R(z,u))]\\ \forall y \forall z[(R(a,y) \wedge R(a,z)) \to \exists u(R(y,u) \wedge R(z,u))]\\ \forall x \forall y \forall z[(R(x,y) \wedge R(x,z)) \to \exists u(R(y,u) \wedge R(z,u))]\\ } $

Is this proof correct ? Am I allowed to apply $\mathbf{\forall E}$ twice on lines 3,4 ?

Answer 1

Strictly, NO, not in a standard Fitch-style system. Look again at the official statement of the $\forall$E rule. You instantiate one variable at a time. Hence it takes two steps to get line (3).

Applying $\forall$E to the premiss once gets you to e.g.

$\forall y(R(a, y) \to R(y, a))$

And then another step takes you to the desired

$(R(a, b) \to R(b, a))$.

Similarly it takes two steps to get line (4).