(New page: == The Problem == Consider the following grammar: <text> bexpr -> bexpr or bexpr | bterm bterm -> bterm and bterm | bfactor bfactor -> not bfactor | ( bexpr ) | true | false </text>...) |
(→The Problem) |
||
Line 3: | Line 3: | ||
Consider the following grammar: | Consider the following grammar: | ||
<text> | <text> | ||
− | bexpr | + | bexpr -> bexpr or bexpr | bterm |
− | bterm | + | bterm -> bterm and bterm | bfactor |
− | bfactor | + | bfactor -> not bfactor | ( bexpr ) | true | false |
</text> | </text> | ||
Consider the following grammar: <text> bexpr -> bexpr or bexpr | bterm bterm -> bterm and bterm | bfactor bfactor -> not bfactor | ( bexpr ) | true | false </text>