we would assert that BUY is subsumed by and(all(giving1,GIVE),all**(giving2,GIVE))**,
and then add further constraints about the attributes of BUY (such as buyer and **seller**)
and those of GIVE (such as giver and receiver), including same-as**([buyer],[giving1**
giver]) and same-as( [seller],**[giving1 receiver]**).

The procedures that need to be written to extend the implementation are: same-as::-
Recognize, same-as::InferFromFilling, same-as::InferFromClosing, same-as::-
InferFromAsserting, same-as::ConsistentWFilling?, same-as::ConsistentW Closing?, and same-as::ConsistentWAsserting?**. In addition, some previously written procedures may need to be augmented.**
Rules of inference for −→ and 6−→ **can be presented as a way of describing the tasks**
performed by the above functions. However, in this case, the **functions also need to worry**
about dependency pointers, so the mapping is less direct.

The following inference rule describes the standard recognition procedure for individuals
(there are additional rules for subsumption of course):

$KB\dashb\dashrightarrow$ fills($[f_{1},\dashrightarrow,f_{n}],e$)  $KB\dashb\dashrightarrow$ fills($[g_{1},\dashrightarrow,g_{m}],e$)  $b\dashrightarrow$ same-as($[f_{1},\dashrightarrow,f_{n}],[g_{1},\dashrightarrow]$
KB ⊢ b −→ same-as([f1, · · ·, fn], [g1, · · ·, gm])
If the recognition function can only follow a chain up to an individual e**, which is missing**
the filler of the next attribute in the chain, then a dependency from e **must be recorded.**
same-as::Recognizes?(b, same-as**(chain1,chain2))**
{ **(e1,p') := follow chain1 from b, returning the furthest**
individual e1 reached, and the remainder of the chain p' which was not possible to follow; if not(p'=nil)
then { post dependency e1 −−−−−−−−−−−−→ RecognizeDependsOnMe **b; return false;**}
(e2,q') := follow chain2 from b, as above; if not(q'=nil)
then { post dependency e2 −−−−−−−−−−−−→ RecognizeDependsOnMe **b; return false;**}
if (e1=e2) & p'=q' then return true else return false.

}
This function will be called from inside and::Recognizes?(b,C) **as follows:**
...

for every same-as(chn1,chn2) in get **sameas(C)**
if not same-as::Subsumes?(same-as**(chn1,chn2), Descriptor(b))**
then if not same-as::Recognizes?(b,same-as**(chn1,chn2))**
then return false.

...

Reasoning with same-as **concerning individuals is actually more complex, because there**
are additional rule of inference, such as KB ⊢ b −→ fills(chain1, e)
KB ⊢ b −→ same-as(chain1**, chain**2)

$$\frac{K B\,\vdash e\longrightarrow\mathbf{same\mbox{-}as(}c h a n3,c h a n4)}{K B\,\vdash b\longrightarrow\mathbf{same\mbox{-}as(}c h a i n1\circ c h a i n3,c h a i n2\circ c h a i n4)}$$
