Theorem dfsb3

Description: An alternate definition of proper substitution df-sb that uses only primitive connectives (no defined terms) on the right-hand side. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 6-Mar-2007) (New usage is discouraged.)

	Ref	Expression
Assertion	dfsb3	$$⊢\left[y/x\right]\phi \leftrightarrow \left(\left(x=y\to \neg \phi \right)\to \forall x\left(x=y\to \phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-or	$$⊢\left(\left(x=y\wedge \phi \right)\vee \forall x\left(x=y\to \phi \right)\right)\leftrightarrow \left(\neg \left(x=y\wedge \phi \right)\to \forall x\left(x=y\to \phi \right)\right)$$
2		dfsb2	$$⊢\left[y/x\right]\phi \leftrightarrow \left(\left(x=y\wedge \phi \right)\vee \forall x\left(x=y\to \phi \right)\right)$$
3		imnan	$$⊢\left(x=y\to \neg \phi \right)\leftrightarrow \neg \left(x=y\wedge \phi \right)$$
4	3	imbi1i	$$⊢\left(\left(x=y\to \neg \phi \right)\to \forall x\left(x=y\to \phi \right)\right)\leftrightarrow \left(\neg \left(x=y\wedge \phi \right)\to \forall x\left(x=y\to \phi \right)\right)$$
5	1 2 4	3bitr4i	$$⊢\left[y/x\right]\phi \leftrightarrow \left(\left(x=y\to \neg \phi \right)\to \forall x\left(x=y\to \phi \right)\right)$$