Theorem sbcom3vv 2132

Description: Substituting 𝑦 for 𝑥 and then 𝑧 for 𝑦 is equivalent to substituting 𝑧 for both 𝑥 and 𝑦. Version of sbcom3 2538 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.)

Assertion

Ref	Expression
sbcom3vv	⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)

Distinct variable group:   𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem sbcom3vv

Step	Hyp	Ref	Expression
1		sbequ 2117	. 2 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
2	1	sbbiiev 2127	1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)

Syntax hints:   ↔ wb 209  [wsb 2096

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-sb 2097

This theorem is referenced by:  sbievw2  2133  sbcovOLD  2293