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\begin{document}

{\large\bf CS 4995}\hspace{.8in}{\large\bf Notes} (S. Cook and T. Pitassi)
\hfill {\large\bf Fall, 2022}

\begin{center}
{\Large\bf Completeness of System $LK$ for Predicate Calculus}\\
\end{center}

\bigskip

In general in this section of the notes we assume that every formula
$A$ satisfies the restriction described on page 27:
All free variables of $A$ are from the free variable list $a,b,c,...$ and all
bound variables of $A$ are from the bound variable list $x,y,z,...$.

On page 29 we proved the soundness of $LK$: Every sequent provable
in $LK$ is valid.   Now we want to prove completeness, which is the
converse:  Every valid sequent is provable in $LK$.  This completeness
theorem, originally proved by \Godel\ in 1930 (for a different proof
system), ranks as one of the
great results in logic of this century.  The definition of a valid formula
(page 23)
is a formula that is true under all interpretations (structures), 
including those with an infinite universe (even an uncountably infinite
universe).  The completeness theorem states that for every valid
formula $A$, this universal truth can be established by exhibiting a finite
syntactic object.  In our case, that object is an $LK$ proof of
the sequent $\ra A$.

Actually we will
prove a stronger result, a kind of derivational completeness, in order
to get the first-order compactness theorem as a corollary.
As in the propositional system $PK$, we want to generalize $LK$ proofs
to allow nonlogical axioms.  We will restrict attention to the simple
case in which every nonlogical axiom has the form $\ra A$, where $A$
is a formula.

%{\bf Notation:}  Let $\Phi$ be a set of formulas.  Then $\ra\Phi$
%is the set of all sequents of the form $\ra A$, where $A$ is in $\Phi$.

{\bf Definition:}  If $\Phi$ is a set of formulas, then an
$LK-\Phi$ proof is an $LK$ proof in which sequents at the leaves may
be either logical axioms of the form $A\ra A$ (for any formula $A$), or
nonlogical axioms of the form $\ra A$, where $A$ is in $\Phi$.

%Notice that a structure $\calM$ satisfies $\ra\Phi$ iff $\calM$
%satisfies $\Phi$.  Also a sequent $\Gamma\ra\Delta$ is a logical
%consequence of $\ra\Phi$ iff $\Gamma\ra\Delta$ is a logical
%consequence of $\Phi$.

We would like to be able to say that a sequent $\Gamma\ra\Delta$
is a logical consequence of a set $\Phi$ of formulas iff there is
an $LK-\Phi$ proof of $\Gamma\ra\Delta$.  Unfortunately the soundness
direction of the assertion is false.  For example, using the
$\forall$ {\bf right} rule we can derive $\ra\forall x Px$ from
$\ra Pb$, but $\ra\forall x Px$ is not a logical consequence of $Pb$.

We could correct the soundness statement by asserting it true for
sentences, but we want to generalize this a little by introducing
the notion of the universal closure of a formula or sequent.

Recall the definition (page 28) of the universal closure $\forall A$
of a formula $A$.  Also if
$\Phi$ is a set of formulas, then $\forall \Phi$ is the set of all
sentences $\forall A$, for $A$ in $\Phi$.
Notice that if $A$ is a sentence (i.e. no free variables),
then $\forall A$ is the same as $A$.

\begin{exer}
\label{closure}
Prove from the definition of logical consequence (page 23) that
every formula $A$ is a logical consequence of its universal
closure $\forall A$.  (I.e. show that $\forall A\models A$.)
Give an example of a formula $A$ such that $A\not\models \forall x A$.
Show however that for every formula $A$ we have $\models A$ iff
$\models \forall A$.
\end{exer}

\begin{exer}
\label{closureImp}
Show that if $ \Phi\models A$ then $\forall \Phi\models \forall A$.
Give an example to show that the converse may be false:  We could
have $\forall \Phi\models \forall A$ but not  $ \Phi\models A$.
\end{exer}

Initially we study the case in which the underlying vocabulary does
not contain =.  To handle the case in which = occurs we must introduce
equality axioms.  This will be done later.

{\bf Theorem:} (Derivational Soundness and Completeness of $LK$)
Assume that the underlying vocabulary does not contain =.
Let $\Phi$ be a set of sequents and let $\Gamma\ra\Delta$ be a
sequent.  Then there is an $LK-\Phi$ proof of $\Gamma\ra\Delta$
iff $\Gamma\ra\Delta$ is a logical consequence of $\forall\Phi$.
The soundness (only if) direction holds also when the
underlying vocabulary contains =.

{\bf Proof of Soundness:}  Let $\pi$ be a $LK-\Phi$
proof of $\Gamma\ra\Delta$.  We must show that $\Gamma\ra\Delta$
is a logical consequence of $\forall\Phi$.  We want to prove this by
induction on the number of sequents in the proof $\pi$, but in
fact we need a stronger induction hypothesis, to the effect that
the ``closure'' of $\Gamma\ra\Delta$ is a logical consequence of
$\forall\Phi$.  So we first have to define the closure of a
sequent.

Thus we define the closure $\forall S$ of a sequent $S$ to be the
closure of its associated formula (page 28):
$\forall S \eqsyn \forall A_S$,
where $A_S \eqsyn \bigwedge\Gamma \supset \bigvee \Delta$, when
$S\eqsyn \Gamma\ra\Delta$.  Note that $\forall S$ is {\em not}
equivalent to $\forall\Gamma\ra\forall\Delta$ in general. 

We now prove by induction on the number of sequents in $\pi$, that
if $\pi$ is an $LK-\Phi$ proof of a sequent $S$, then $\forall S$
is a logical consequence of $\forall\Phi$.  It will follow by Exercise
\ref{closure} above that $S$ itself is a logical consequence of $\forall\Phi$,
and so Soundness follows.

For the base case, the sequent $S$ is
either a logical axiom $A\ra A$, which is
valid and hence a consequence of $\forall\Phi$, or it is a nonlogical axiom
$\ra A$, where $A$ is a formula in $\Phi$.  In the latter case,
$\forall S$ is equivalent to $\forall A$, which of course is a logical
consequence of $\forall\Phi$.

For the induction step, it is sufficient to check that for each rule
of $LK$, the closure of the bottom sequent is a logical consequence of the
closure(s) of the sequent(s) on top.   With two exceptions, this
statement is true when the word ``closure'' is omitted, and adding
back the word ``closure'' then follows from Exercise \ref{closureImp}
above.  The two exceptions are the
rules, $\forall$-{\bf right} and $\exists$-{\bf left}.  For these, the bottom
is {\em not} a logical consequence of the top in general, but 
it is not hard to show that the closures of the top and bottom are
equivalent.  Note however that the {\bf Restriction} on the free variable
$b$ in these rules (page 28) is essential in order to show
that the closure of the bottom is a logical consequence of the
closure of the top.  $\Box$

The proof of completeness is more difficult and more interesting than
the proof of soundness.  The basic idea is similar to the proof for
the propositional system PK, but unfortunately the Inversion Principle
fails for two of the new LK rules:  $\exists$-{\bf right} and
$\forall$-{\bf left}.  To get around this we will have to apply
these rules repeatedly using different terms $t$, and
the contraction rule becomes necessary (at least for cut-free proofs).

The following lemma lies at the heart of
this proof.

{\bf Completeness Lemma:}  Assume that the underlying vocabulary does
not contain =.  If $\Gamma\ra\Delta$ is a sequent and $\Phi$ is a
set of formulas such that $\Gamma\ra\Delta$ is a
logical consequence of $\Phi$, then there is a finite subset
$\{C_1,...,C_n\}$ of $\Phi$ such that the sequent
$$      C_1,..., C_n,\Gamma\ra \Delta $$
has an $LK$ proof $\pi$ which does not use the cut rule.

{\bf Proof of Derivational Completeness from the Completeness Lemma:}
\label{DCCLemma}
Let $\Phi$ be a set of formulas such
that $\Gamma\ra\Delta$ is a logical consequence of $\forall\Phi$.
By the completeness lemma, there is a finite subset
$\{C_1,...,C_n\}$ of $\Phi$ such that
$$\forall C_1,..., \forall C_n,\Gamma\ra \Delta $$
has a cut-free $LK$ proof $\pi$.
Note that for each $i, 1\leq i\leq n$, the sequent $\ra \forall C_i$
has an $LK-\Phi$ proof from the nonlogical axiom $\ra C_i$ by repeated
use of the rule $\forall$-{\bf right}.  Now the proof $\pi$ can
be extended, using these proofs of the sequents 
$$      \ra\forall C_1 \ \ ... \ra\forall C_n  $$
and repeated use of the cut rule, to form an $LK-\Phi$ proof
$\Gamma\ra\Delta$.  $\Box$

{\bf Proof of the Completeness Lemma:}  We loosely follow the proof of the
Cut-free Completeness Theorem, pp 33-36 of [Buss].  (Warning:  our
definition of logical consequence differs from Buss's when the
formulas in the hypotheses have free variables.)  We will only prove it for
the case in which the underlying first-order vocabulary $\SL$ 
has a countable set (including the case of a finite set)
of function and predicate symbols; i.e.
the function symbols form a list $f_1,f_2,...$ and the predicate
symbols form a list $P_1,P_2,...$.  This may not seem like much
of a restriction, but for example in developing the model theory
of the real numbers, it is sometimes useful to introduce a distinct
constant symbol $e_c$ for every real number $c$; and there are
uncountably many real numbers.  The completeness theorem and lemma
hold for the uncountable case, but we shall not prove them for
this case.

For the countable case, we may assign a distinct binary string to
each function symbol, predicate symbol, variable, etc. and hence
assign a unique binary string to each formula and term.  This allows
us to enumerate all the $\SL$-formulas in a list $A_1,A_2,...$
and enumerate all the $\SL$-terms in a list $t_1,t_2,...$.  Here
we assume that all free variables in each formula $A_i$ or term
$t_j$ come from the list $a,b,c,...$ of free variables.  Further
we may assume that every formula occurs infinitely often in the
list of formulas, and every term occurs infinitely often in the list
of terms.  (For example, we can take the original sequence of terms
and form a new sequence $t_1,t_1,t_2,t_1,t_2,t_3,...$ which satisfies
this condition.)  Finally we may enumerate all pairs $\langle A_i,t_j\rangle$,
using any method of enumerating all pairs of natural numbers.

We are trying to find an $LK$ proof of some
sequent of the form $C_1,...,C_n, \Gamma\ra\Delta$, for some $n$.
Starting with $\Gamma\ra\Delta$ at the bottom, we work upwards by
applying the rules in reverse, much as
in the proof of the propositional completeness theorem for $PK$.
However now we will add formulas $C_i$ to the antecedent from time
to time.  Also unlike the $PK$ case we have no inversion principle
to work with (specifically for the rules $\forall$-{\bf left}
and $\exists$-{\bf right}).  Thus it may happen that our proof-building
procedure may not terminate.  In this case we will show
how to define a structure which shows that $\Gamma\ra\Delta$ is not
a logical consequence of $\Phi$.

We construct our cut-free proof tree $\pi$ in stages.  Initially
$\pi$ consists of just the sequent $\Gamma\ra\Delta$.  At each
stage we modify $\pi$ by possibly adding a formula from $\Phi$ to the
antecedent of every sequent in $\pi$, and by adding
subtrees to some of the leaves.

{\bf Definition:}\label{activeS}
A sequent in $\pi$ is said to be {\em active} provided it is at
a leaf, and no formula occurs in both its antecedent and succedent.

Note that if some formula does occur on both sides, the sequent
can be derived from a logical axiom with weakenings and exchanges.

Each stage uses one pair in our enumeration of all pairs
$\langle A_i,t_j\rangle$.  Here is the procedure for the
next stage, in general.

==================================================

Let $\langle A_i,t_j\rangle$ be the next pair in the enumeration.
We call $A_i$ the {\em active} formula for this stage.

\label{stepOne}
Step (1):  If $A_i$ is in $\Phi$, then replace every sequent
$\Gamma'\ra\Delta'$ in $\pi$ with the sequent $\Gamma',A_i\ra\Delta'$.

Step (2):  If $A_i$ is atomic, do nothing and proceed to the next stage.
Otherwise, modify $\pi$ at the active sequents which contain $A_i$
by applying the appropriate introduction rule in reverse, much as
in the propositional completeness proof on page 13.  For example,
if $A_i$ is of the form $B\vee C$, then every active sequent in $\pi$
of the form $\Gamma',B\vee C,\Gamma''\ra\Delta'$ is replaced by the
derivation
$$
\begin{prooftree} \label{orLeft}
 \Gamma',B,\Gamma''\ra\Delta'  \qquad
        \Gamma',C,\Gamma''\ra\Delta' 
   \Justifies \Gamma',B\vee C,\Gamma''\ra\Delta'
\end{prooftree}
$$
Here the double line represents a derivation involving the rule
$\vee$-{\bf left}, together with exchanges to move the principle
formulas to the left end of the antecedent and back.  The treatment
is similar for the other propositional cases.

If $A_i$ is of the form $\exists xB(x)$, then every active sequent of $\pi$
of the form $\Gamma',\exists xB(x),\Gamma''\ra\Delta'$ is replaced by the
derivation 
$$
\begin{prooftree}
 \Gamma',B(c),\Gamma''\ra\Delta'
   \Justifies \Gamma',\exists xB(x),\Gamma''\ra\Delta'
\end{prooftree}
$$
\label{cRestr}
where $c$ is a new free variable, not used in $\pi$ yet.  (Also
$c$ may not occur in any formula in $\Phi$, because otherwise at a
later stage, Step (1) of the procedure might cause the variable restriction
in the $\exists$-{\bf left} rule to be violated.) 

In addition,
any active sequent of the form $\Gamma'\ra\Delta',\exists xB(x),\Delta''$
is replaced by the derivation
$$
\begin{prooftree}\label{existsCase}
 \Gamma'\ra\Delta',\exists xB(x),B(t_j),\Delta''
   \Justifies  \Gamma'\ra\Delta',\exists xB(x),\Delta''
\end{prooftree}
$$
Here the term $t_j$ is the second component in the current pair
$\langle A_i,t_j\rangle$.  The derivation uses the rule
$\exists$-{\bf right} to introduce a new copy of $\exists xB(x)$,
and then the rule contraction-{\bf right} to combine the two
copies of $\exists xB(x)$. 

NOTE: This and the dual $\forall$-{\bf left}
case are the only two cases that use the term $t_j$, and the only
cases that use the {\bf contraction rules}.

The case where $A_i$ begins with a universal quantifier is dual to
the above existential case.

Step (3)  If there are no active sequents remaining in $\pi$, then
exit from the algorithm.  Otherwise continue to the next stage.

================================================================

\begin{exer}
Carry out the case above in which $A_i$ begins with a universal quantifier.
\end{exer}

If the algorithm constructing $\pi$ ever halts, then $\pi$ gives
a cut-free proof of $\Gamma,C_1,...,C_n\ra\Delta$ for some formulas
$C_1,...,C_n$ in $\Phi$.  This is because the nonactive leaf sequents
all contain a formula $A$ which occurs on both the left and right,
and each such sequent can be derived from the logical axiom $A\ra A$,
using weakenings and exchanges.  Thus $\pi$ can be extended,
using exchanges, to a cut-free proof of $C_1,...,C_n,\Gamma\ra\Delta$,
as desired.

It remains to show that if the above algorithm constructing $\pi$
never halts, then the sequent $\Gamma\ra\Delta$ is not a logical
consequence of $\Phi$.  So suppose the algorithm never halts, and
let $\pi$ be the result of running the algorithm forever.  In general,
$\pi$ will be an infinite tree, although in special cases $\pi$ is
a finite tree.  In general the objects at the leaves of the tree
will not be finite sequents, but because of Step (1) of the algorithm
above, they will be 
of the form $\Gamma',C_1,C_2,...\ra\Delta'$, where $C_1,C_2,...$
is an infinite sequence of formulas containing all formulas in $\Phi$,
each repeated infinitely often (unless $\Phi$ is empty).  We shall
refer to these infinite pseudo-sequents as just ``sequents''.

If $\pi$ has only finitely many nodes, then at least one leaf node
must be active (and contain only atomic formulas), since
otherwise the algorithm would terminate.  In
this case, let $\beta$ be a path in $\pi$ from the root extending
up to this active node.  If on the other hand $\pi$ has infinitely
many nodes, then by \Konig's Lemma (page 16),  there must be an
infinite branch $\beta$ in $\pi$ starting at the root and extending up
through the tree.  Thus in either case, $\beta$ is a branch in $\pi$
starting at the root, extending up through the tree, and such that
all sequents on $\beta$ were once active, and hence have no formula occurring
on both the left and right.

{\bf Term model construction}
\label{termModel}

We use this branch $\beta$ to construct a structure $\calM$ and
an object assignment $\sigma$ which satisfy every formula in $\Phi$,
but falsify the sequent $\Gamma\ra\Delta$ (so $\Gamma\ra\Delta$ is
not a logical consequence of $\Phi$).

The universe $M$ of $\calM$
is the set of all $\SL$-terms $t$ (which contain only ``free'' variables
$a,b,c,...$).
(Hence $\calM$ is called a {\em term model}.)

The object assignment $\sigma$ just maps every variable
$a$ to itself.  

The interpretation $f^{\calM}$ of each $k$-ary function symbol $f$
is defined so that $f^{\calM}(r_1,...,r_k)$ is the term $fr_1...r_k$,
where $r_1,...,r_k$ are any terms (i.e. any members of the universe).

The interpretation $P^{\calM}$ of each $k$-ary predicate symbol $P$
is defined by letting $P^{\calM}(r_1,...,r_k)$ hold iff the atomic
formula $Pr_1...r_k$ occurs in the antecedent (left side) of some sequent
in the branch $\beta$.

\begin{exer}
\label{term}
Prove by structural induction that for every term $t$, $t^{\calM}[\sigma] = t$.
\end{exer}

{\bf Claim:}\label{Formclaim}
For every formula $A$, if $A$ occurs in some antecedent
in the branch $\beta$, then $\calM$ and $\sigma$ satisfy $A$, and
if $A$ occurs in some succedent in $\beta$, then  $\calM$ and $\sigma$
falsify $A$.

Since the root of $\pi$ is the sequent $\Gamma,C_1,C_2,...\ra\Delta$,
where $C_1,C_2,...$ contains all formulas in $\Phi$, it follows that
$\calM$ and $\sigma$ satisfy $\Phi$ and falsify $\Gamma\ra\Delta$.

We prove the Claim by structural induction on formulas $A$.  For the
base case, if $A$
is an atomic formula, then by the definition of $P^{\calM}$ above,
$A$ is satisfied iff $A$ occurs in some antecedent of $\beta$.
But no atomic formula can occur both in an antecedent of some node
in $\beta$
and in a succedent (of possibly some other node) in  $\beta$,
since then these formulas would
persist upward in $\beta$ so that some particular sequent
in $\beta$ would have $A$ occurring both on the left and on the right.
Thus if $A$ occurs in some succedent of $\beta$, it
is not satisfied by $\calM$ and $\sigma$.

For the induction step, there is a different case for each of the
ways of constructing a formula from simpler formulas (see the 
definition of formula, page 19).  In general, if $A$ occurs in some
sequent in $\beta$, then $A$ persists upward in every higher sequent
of $\beta$ until it becomes the active formula ($A\eqsyn A_i$).
Each case is handled by the
corresponding introduction rule used in the algorithm.
For example, if $A$ is of the form $B\vee C$ and $A$ occurs on
the left of a sequent in $\beta$, then the rule $\vee$-{\bf left}
is applied in reverse (see page \pageref{orLeft}),
so that when $\beta$ is extended upward
either it will have some antecedent containing $B$ or one containing $C$.
In the case of $B$, we know that $\calM$ and $\sigma$ satisfy $B$ by
the induction hypothesis, and hence they satisfy $B\vee C$.
(Similarly for $C$.)  

Now consider the interesting case in which $A$ is $\exists xB(x)$ and $A$
occurs in some succedent of $\beta$.  See page \pageref{existsCase}
to find out what happens when $A$ becomes active.  The path 
$\beta$ will hit a sequent with $B(t_j)$ in the succedent,
and by the induction hypothesis, $\calM$ and $\sigma$ falsify 
$B(t_j)$.
But this succedent still has a copy of $\exists xB(x)$, and in fact
this copy will be in {\em every} succedent of $\beta$ above this
point.  Hence {\em every} $\SL$-term $t$ will eventually be of the
form $t_j$ and so the formula $B(t)$ will occur as a succedent on
$\beta$.  (This is why we assumed that every term appears infinitely
often in the sequence $t_1,t_2,...$.)  Therefore $\calM$ and $\sigma$
falsify $B(t)$ for every term $t$ (i.e. for every element in the universe
of $\calM$).  Therefore they falsify $\exists xB(x)$, as required.

This and the dual case in which $A$ is $\forall xB(x)$ and occurs
in some antecedent of $\beta$ are the only subtle cases.  All other
cases are straightforward.  $\Box$.

\begin{exer}
Consider the sequent
$$
    Pe, \forall x(\neg Px\vee Pffx)\ra Pfe
$$
where $e$ is a constant and $f$ is a unary function symbol.
This sequent is not valid, and hence the procedure described in the
proof of the Completeness Lemma will not terminate with an LK proof.
In this case it is sufficient to consider the set
$$
   T = \{e, fe, ffe, fffe, \ldots \}
$$
of terms.  Carry out the construction (repeatedly apply Step (2))
in the proof to get an LK tree $\pi$ with
an infinite branch, where the terms used in
the $\forall$-{\bf Left} rule are restricted to the set $T$.  
(You do not need to describe all of $\pi$, but describe one infinite
path in $\pi$ and describe the resulting term model.  Show that it
falsifies the sequent.)
\end{exer}

\begin{exer}
\label{completeness}
What goes wrong in the above proof if $\Gamma\ra\Delta$ contains =?
\end{exer}

\begin{exer}
\label{contr-rule}
Show how to use a contraction rule to get a cut-free proof of
the sequent
$$
    \forall x(\neg Px \vee Pfx) \ra \forall x(\neg Px \vee Pffx)
$$
\end{exer}

We have already proved derivational completeness from the
Completeness Lemma (page \pageref{DCCLemma}),
but now we mention a stronger
form of derivational completeness ({\em anchored completeness})
which shows that the cut formulas in the LK derivation can
be restricted to formulas in the hypothesis set $\Phi$, provided
that $\Phi$ is closed under substitution of terms for variables
(i.e., if $A(b)$ is in $\Phi$, and $t$ is a proper term (page 27)
then $A(t)$ is also in $\Phi$.)  Note that if $\Phi$ is a set
of sentences, it is automatically closed under substitution of
terms for variables.
 
{\bf Anchored Completeness Theorem:}
Assume that the underlying vocabulary is countable
and does not contain =.
Suppose that $\Phi$ is a set of formulas closed under substitution
of terms for variables. 
Suppose that $\Gamma\ra\Delta$ is a sequent that is a logical consequence
of $\forall \Phi$.  Then there is an $LK-\Phi$ proof of $\Gamma\ra\Delta$
in which the cut rule is restricted so that the only cut formulas
are formulas in $\Phi$.
 
Note that if all formulas in $\Phi$ are sentences, then the above
theorem follows easily from the Completeness Lemma, since in this
case $\forall\Phi$ is the same as $\Phi$.  However if formulas
in $\Phi$ have free variables, then apparently the cut rule must
be applied to the closures $\forall C$ of formulas in $\Phi$ (as
opposed to $C$ itself) in
order to get an $LK-\Phi$ proof of $\Gamma\ra\Delta$, as explained
on page \pageref{DCCLemma}.  To see that the cut formulas can be taken
to be in $\Phi$ instead of closures of formulas in $\Phi$,
the proof of the Completeness Lemma can be modified, as explained
in the following exercise.
 
\begin{exer}
Show how to prove the above theorem by modifying the proof of
the Completeness Lemma according to  a),...,d) below.
\end{exer}
a) the definition of {\em active sequent} on page \pageref{activeS} must
be modified, since now we're allowing nonlogical axioms in $\pi$.
Give the precise new definition.
 
b) Step (1) of the procedure on page \pageref{stepOne} must be modified,
because now we're looking for a derivation of $\Gamma\ra\Delta$ from
nonlogical axioms, rather
than a proof of $C_1,...,C_n,\Gamma\ra\Delta$.  Describe the modification.
(We still need to bring formulas $A_i$ of $\Phi$ somehow into the proof,
and your modification will involve adding a short derivation to $\pi$.)
 
c) the restriction given on page \pageref{cRestr} for the case $\exists xB(x)$
on the left,
that the variable $c$ must not occur in any formula in $\Phi$, must
be dropped.  Explain why.
 
d) Explain why the structure $\calM$ and object assignment $\sigma$,
described on page \pageref{termModel}, satisfy $\forall \Phi$. 
This should follow
from the Claim on page \pageref{Formclaim}, and your modification of Step (1),
which should ensure that each formula in $\Phi$ occurs in the
antecedent of some sequent in every branch in $\pi$.
Conclude that $\Gamma\ra\Delta$
is not a logical consequence of $\forall\Phi$ (when the procedure
does not terminate).






\end{document}
