Theorem sbcom3vv

Description: Substituting y for x and then z for y is equivalent to substituting z for both x and y . Version of sbcom3 with a disjoint variable condition using fewer axioms. (Contributed by NM, 27-May-1997) (Revised by Giovanni Mascellani, 8-Apr-2018) (Revised by BJ, 30-Dec-2020) (Proof shortened by Wolf Lammen, 19-Jan-2023)

	Ref	Expression
Assertion	sbcom3vv	$$⊢\left[z/y\right]\left[y/x\right]\phi \leftrightarrow \left[z/y\right]\left[z/x\right]\phi$$

Proof

Step	Hyp	Ref	Expression
1		sbequ	$$⊢y=z\to \left(\left[y/x\right]\phi \leftrightarrow \left[z/x\right]\phi \right)$$
2	1	pm5.74i	$$⊢\left(y=z\to \left[y/x\right]\phi \right)\leftrightarrow \left(y=z\to \left[z/x\right]\phi \right)$$
3	2	albii	$$⊢\forall y\left(y=z\to \left[y/x\right]\phi \right)\leftrightarrow \forall y\left(y=z\to \left[z/x\right]\phi \right)$$
4		sb6	$$⊢\left[z/y\right]\left[y/x\right]\phi \leftrightarrow \forall y\left(y=z\to \left[y/x\right]\phi \right)$$
5		sb6	$$⊢\left[z/y\right]\left[z/x\right]\phi \leftrightarrow \forall y\left(y=z\to \left[z/x\right]\phi \right)$$
6	3 4 5	3bitr4i	$$⊢\left[z/y\right]\left[y/x\right]\phi \leftrightarrow \left[z/y\right]\left[z/x\right]\phi$$