Example 2. Consider the following proposition A1VA2∧A which is definitely not well-defined Solution. We have two possible different propositions 1.(A1VA2)∧A3 2.A1V(A2∧A3) Of course, they have different abbreviation truth tables Mathematics is rigid. One of the responsibilities of mathematical logic is to eliminate ambiguity from mathematics. So we should first eliminate ambiguity in mathematical logic 2.2 Parentheses Finally, we want to design a systematic approach to recognize/parse a well defined proposition But we first dig some properties of well defined propositions A proposition is a sequence of symbols generated by proposition language. We just take it string. Parentheses are very important to construct a well-defined proposition. We just count number of left and right parenthesis and have the following theorem Theorem 1. Every well-defined proposition has the same number of left as right parentheses Proof. Consider the symbols without parenthesis first. They are the simplest cases: propositional letters. And we know that a complicated proposition consists of two or one simpler proposition(s) with a connective. If a simpler one keep the property, we can derive the invariant for the complicated Obviously, we can prove it by induction to investigate all propositions by following the approach to construct a compound proposition What metric is chosen for induction? It is magic. Here, we actually use the depth of nested connectives, which will be introduced in the next section. Some guy may take the number of parenthesis. It is difficult to characterize the feature and to apply inductive proof. The exercise will demonstrate this Theorem 2. Any proper initial segement of a well defined proposition contains an excess of left trenthesiss. Thus no proper initial segement of a well defined propositon can itself be a well defined propositions Go back to lecture three if your forget what is a proper initial segment Proof. Prove it by induction from simple to complicated propositions. And we also need Theorem 1Example 2. Consider the following proposition A1 ∨ A2 ∧ A3, which is definitely not well-defined. Solution. We have two possible different propositions 1. (A1 ∨ A2) ∧ A3 2. A1 ∨ (A2 ∧ A3) Of course, they have different abbreviation truth tables. Mathematics is rigid. One of the responsibilities of mathematical logic is to eliminate ambiguity from mathematics. So we should first eliminate ambiguity in mathematical logic. 2.2 Parentheses Finally, we want to design a systematic approach to recognize/parse a well defined proposition. But we first dig some properties of well defined propositions. A proposition is a sequence of symbols generated by proposition language. We just take it as a string. Parentheses are very important to construct a well-defined proposition. We just count the number of left and right parenthesis and have the following theorem. Theorem 1. Every well-defined proposition has the same number of left as right parentheses. Proof. Consider the symbols without parenthesis first. They are the simplest cases: propositional letters. And we know that a complicated proposition consists of two or one simpler proposition(s) with a connective. If a simpler one keep the property, we can derive the invariant for the complicated one. Obviously, we can prove it by induction to investigate all propositions by following the approach to construct a compound proposition. What metric is chosen for induction? It is magic. Here, we actually use the depth of nested connectives, which will be introduced in the next section. Some guy may take the number of parenthesis. It is difficult to characterize the feature and to apply inductive proof. The exercise will demonstrate this. Theorem 2. Any proper initial segement of a well defined proposition contains an excess of left parenthesiss. Thus no proper initial segement of a well defined propositon can itself be a well defined propositions. Go back to lecture three if your forget what is a proper initial segment. Proof. Prove it by induction from simple to complicated propositions. And we also need Theorem 1. 2