proof

Case 1: k+1 is a prime. In this case k+1=k+1k+1 = k+1 shows that P(k+1)≡TP( k+1 ) \equiv T

Define predicate P(n)P( n ) as the statement that nn equals to the product of some primes, where the domain of nn is {n∈Z|n>1}\{n \in Z | n > 1\} so the smallest element in this domain is 22.

Case 2: k+1 is a not a prime. In this case k+1=a×bk+1 = a \times b , where aa and bb are positive integers greater than 1.

Using the direct proof technique, we assume that (P(2)∧⋯∧P(k))≡T(P(2) \wedge \cdots \wedge P( k )) \equiv T.

QED

To use the UG rule, we consider an arbitrary positive integer kk with k≥2k \geq 2.

Inductive step: to prove that ∀k((P(2)∧⋯∧P(k))→P(k+1))≡T\forall k ( (P(2) \wedge \cdots \wedge P( k )) \to P( k+1 ) ) \equiv T, where k≥2k \geq 2.

Basis step: to prove that P(2)≡TP(2) \equiv T.

Since 2 is a prime, 2=22=2 shows that P(2)≡TP(2) \equiv T.

So k+1=a×bk+1 = a\times b can be expressed by the product of some primes, i.e., P(k+1)≡TP( k+1 ) \equiv T

For P(k+1)P( k+1 ), there are two cases.

Therefore, (P(a)≡T)∧(P(b)≡T)(P( a ) \equiv T) \wedge (P( b ) \equiv T), which means that aa and bb can both be expressed by the product of some primes.

Basis step: to prove that P(1)≡TP( 1 ) \equiv T.

Define predicate P(n)P( n ) as the equation n∑i=11i×(i+1)=nn+1\displaystyle \sum_{i=1}^n \frac{1}{i\times(i+1)}= \frac{n}{n+1}, where the domain of nn is Z+Z^+ and the smallest element in the domain is 1.

Using the direct proof technique, we assume that P(k)≡TP( k ) \equiv T so k∑i=11i×(i+1)=k(k+1)\displaystyle \sum_{i=1}^k \frac{1}{i\times(i+1)}= \frac{k}{(k+1)}.

QED

For P(k+1)P( k+1 ), the LHS =k+1∑i=11i×(i+1)=k∑i=11i×(i+1)+1(k+1)×(k+2)=k(k+1)+1(k+1)×(k+2)=\displaystyle \sum_{i=1}^{k+1} \frac{1}{i\times(i+1)}= \sum_{i=1}^{k} \frac{1}{i\times(i+1)} + \frac{1}{(k+1)\times(k+2)} =  \frac{k}{(k+1)} + \frac{1}{(k+1)\times(k+2)} .

To use the UG rule, we consider an arbitrary positive integer kk.

Inductive step: to prove that ∀k(P(k)→P(k+1))≡T\forall k ( P( k ) \to P( k+1 ) ) \equiv T.

proof

Since k(k+1)+1(k+1)×(k+2)=k×(k+2)+1(k+1)×(k+2)=(k+1)2(k+1)×(k+2)=(k+1)(k+2)\displaystyle \frac{k}{(k+1)} + \frac{1}{(k+1)\times(k+2)} =  \frac{k\times(k+2)+1}{(k+1)\times(k+2)} = \frac{(k+1)^2}{(k+1)\times(k+2)} = \frac{(k+1)}{(k+2)}, P(k+1)≡TP( k+1 ) \equiv T.

For P(1)P( 1 ), the left-hand side (LHS) =1∑i=11i×(i+1)=12==\displaystyle \sum_{i=1}^1 \frac{1}{i\times(i+1)}= \frac{1}{2}= the right-hand side (RHS), so P(1)≡TP( 1 ) \equiv T.