Theorem freq1

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 9-Mar-1997)

	Ref	Expression
Assertion	freq1	$$⊢R=S\to \left(R\mathrm{Fr}A\leftrightarrow S\mathrm{Fr}A\right)$$

Proof

Step	Hyp	Ref	Expression
1		breq	$$⊢R=S\to \left(zRy\leftrightarrow zSy\right)$$
2	1	notbid	$$⊢R=S\to \left(\neg zRy\leftrightarrow \neg zSy\right)$$
3	2	rexralbidv	$$⊢R=S\to \left(\exists y\in x\forall z\in x\neg zRy\leftrightarrow \exists y\in x\forall z\in x\neg zSy\right)$$
4	3	imbi2d	$$⊢R=S\to \left(\left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zRy\right)\leftrightarrow \left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zSy\right)\right)$$
5	4	albidv	$$⊢R=S\to \left(\forall x\left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zRy\right)\leftrightarrow \forall x\left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zSy\right)\right)$$
6		df-fr	$$⊢R\mathrm{Fr}A\leftrightarrow \forall x\left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zRy\right)$$
7		df-fr	$$⊢S\mathrm{Fr}A\leftrightarrow \forall x\left(\left(x\subseteq A\wedge x\ne \varnothing \right)\to \exists y\in x\forall z\in x\neg zSy\right)$$
8	5 6 7	3bitr4g	$$⊢R=S\to \left(R\mathrm{Fr}A\leftrightarrow S\mathrm{Fr}A\right)$$