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!set methtit2=Multiply two positive inequalities
!set methtit=Multiply two inequalities
!set methenv=ZZ QQ RR
!set methparmrelax=3,4
!set methparmtype=parm inequality >,\
parm inequality >,\
parm inequality >,\
parm inequality >
!set methhelp=This method applies the following property. \
<p class="wimscenter" style="font-style:italic">\
If \(A <= B) and \(C <= D) with \(B >= 0) and \(C >= 0), then\
\(AC <= BD).</p>\
The result (\(AC <= BD)) will be added to the hypotheses.\
<p>\
In the case where certains original inequalities are strict, the method\
also knows when one can have \(AC < BD).\
</p>
!if $wims_read_parm iswordof form check
!goto $wims_read_parm
!endif
!exit
:form
One applies the following property.
<p class="wimscenter" style="font-type:italic">If \(A <= B) and \(C <= D) with \(B >= 0) and \(C >= 0), then
\(AC <= BD).
</p>
!set ch_optional=it's obvious
For \(A < B) : the inequality
!read deduc/methparm.phtml 1
where \(B >= 0) because
!read deduc/methparm.phtml 3
<br>
For \(C < D) : the inequality
!read deduc/methparm.phtml 2
where \(C >= 0) because
!read deduc/methparm.phtml 4
!exit
:check
gt=>
lt=<
eq=$empty=
!for i=1 to 4
!distribute items $(methparmobj$i) into data,left$i,sign$i,right$i
mt$i=$(left$i) $(sign$i) $(right$i)
!if $lt isin $(sign$i)
!exchange left$i,right$i
!endif
!next i
methexp=Multiply \($mt1) and \($mt2)
!read deduc/sub/checkineq ($right1),>=,0\
($left1),>=,0\
($right2),>=,0\
($right1),>,0\
($left1),>,0
!distribute lines $out into t1,t2,t3,t4,t5
!if $t2=false
error=Your first inequality does not meet the condition!
!advance penalty
!exit
!endif
!if $t1=true or $t2=true
!if $t2=true
sign3=$sign1=
mt2=$t5
!else
sign3=$sign1
mt2=$t4
!endif
!goto ok1
!endif
!ifval $methparm3=0
!goto bad1
!endif
!read deduc/sub/checkineq ($left1) - ($left3) + ($right3),>=,0\
($left1) - ($left3) + ($right3),>,0
!distribute lines $out into mt1,mt2
!if $mt1!=true
:bad1
error=I don't see why the first inequality meets the condition.
!advance penalty
!exit
!endif
:ok1
!read deduc/sub/checkineq ($right2),>=,0\
($right2),>,0
!distribute lines $out into mt3,mt4
!if $mt3=true
sign4=$sign2=
!goto ok2
!endif
!if $out=false
error=Your second inequality does not meet the condition!
!advance penalty
!exit
!endif
!if $[$methparm4]=0
!goto bad2
!endif
!read deduc/sub/checkineq ($right2) - ($left4) + ($right4),>=,0
($right2) - ($left4) + ($right4),>,0
!distribute lines $out into mt3,mt4
!if $mt3!=true
:bad2
error=I don't see why the second inequality meets the condition.
!exit
!endif
:ok2
!read deduc/sub/simplify ($left1) * ($left2)\
($right1) * ($right2)
!distribute lines $out into newright, newleft
!if $newleft=$empty or $newright=$empty
error=bad_data
!exit
!endif
s_=<
!reset eq1,eq2
!if $eq isin $sign3 and $mt2!=true
eq1=true
!endif
!if $eq isin $sign4 and $mt4!=true
eq2=true
!endif
!if ($eq isin $sign1 and $eq isin $sign2) or\
($eq1=true and $eq2=true) or\
($eq isin $sign1 and $eq1=true) or ($eq isin $sign2 and $eq2=true)
s_=$s_=
!endif
newobject0=$newleft $s_ $newright
oldobject=0
!exit