Searched refs:inequality (Results 1 - 7 of 7) sorted by relevance

/macosx-10.10/Security-57031.1.35/Security/include/security_codesigning/
H A Dreqinterp.h57 bool inequality(CFTypeRef candidate, CFStringCompareFlags flags, CFComparisonResult outcome, bool negate) const;
H A Dreqinterp.cpp502 return inequality(candidate, kCFCompareNumerically, kCFCompareLessThan, true);
504 return inequality(candidate, kCFCompareNumerically, kCFCompareGreaterThan, true);
506 return inequality(candidate, kCFCompareNumerically, kCFCompareGreaterThan, false);
508 return inequality(candidate, kCFCompareNumerically, kCFCompareLessThan, false);
516 bool Requirement::Interpreter::Match::inequality(CFTypeRef candidate, CFStringCompareFlags flags, function in class:Security::CodeSigning::Requirement::Interpreter::Match
/macosx-10.10/Security-57031.1.35/Security/libsecurity_codesigning/lib/
H A Dreqinterp.h57 bool inequality(CFTypeRef candidate, CFStringCompareFlags flags, CFComparisonResult outcome, bool negate) const;
H A Dreqinterp.cpp502 return inequality(candidate, kCFCompareNumerically, kCFCompareLessThan, true);
504 return inequality(candidate, kCFCompareNumerically, kCFCompareGreaterThan, true);
506 return inequality(candidate, kCFCompareNumerically, kCFCompareGreaterThan, false);
508 return inequality(candidate, kCFCompareNumerically, kCFCompareLessThan, false);
516 bool Requirement::Interpreter::Match::inequality(CFTypeRef candidate, CFStringCompareFlags flags, function in class:Security::CodeSigning::Requirement::Interpreter::Match
/macosx-10.10/tcl-105/tcl_ext/tcllib/tcllib/modules/math/
H A Dmachineparameters.tcl145 set inequality [expr {1.0+$epsilon>1.0}]
146 if {$inequality==0} then {
/macosx-10.10/vim-55/runtime/syntax/
H A Dasm68k.vim222 syn match asm68kOperator "<>" " inequality
/macosx-10.10/Heimdal-398.1.2/lib/hcrypto/libtommath/
H A Dtommath.tex3484 this method is faster. Assuming no further recursions occur, the difference can be estimated with the following inequality.
3493 For example, on an AMD Athlon XP processor $p = {1 \over 3}$ and $q = 6$. This implies that the following inequality should hold.
5323 This is trivially true since $x \ge x_s\beta^s$. Next we replace $\hat kx_s\beta^s$ by the previous inequality for $\hat kx_s$.
5329 By simplifying the previous inequality the following inequality is formed.

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