QED

proof

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

Define predicate P(n)P(n)P( n ) with the domain of nn and the smallest element aa in the domain.

Basis step: to prove that P(a)≡TP( a ) \equiv T, where aa is the smallest element in the domain of nn.

QED

proof

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

Define predicate P(n)P( n ) with the domain of nn and the smallest element aa in the domain.

Basis step: to prove that (P(a)∧⋯∧P(b))≡T(P( a ) \wedge \cdots \wedge P(b)) \equiv T, where aa is the smallest element in the domain of nn and b≥ab\geq a is an element in the domain of nn.

proof

Using the EI rule, there is an integer kk such that n=2kn=2k.

QED

n2≡0n^2 \equiv 0 (mod 44)

n2=(2k)2=4k2n^2 = (2k)^2 = 4k^2

Using the direct proof technique, we assume nn is even.

The statement we need to prove is a universally quantified statement, so we consider an arbitrary integer nn.

proof

Using the EI rule, there is an integer kk such that n=2kn=2k.

QED

n2≡0n^2 \equiv 0 (mod 44)

n2=(2k)2=4k2n^2 = (2k)^2 = 4k^2

Using the direct proof technique, we assume nn is even.

The statement we need to prove is a universally quantified statement, so we consider an arbitrary integer nn.