Theorem sbequiOLD 2121

Description: Obsolete version of sbequ 2120 as of 10-Aug-2026, as per conventions 30781, paragraph "Biconditional". (Contributed by NM, 14-May-1993.) (Proof shortened by Wolf Lammen, 15-Sep-2018.) (Proof shortened by Steven Nguyen, 7-Jul-2023.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbequiOLD	⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜑 → [𝑦 / 𝑧]𝜑))

Proof of Theorem sbequiOLD

Step	Hyp	Ref	Expression
1		sbequ 2120	. 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜑))
2	1	biimpd 232	1 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜑 → [𝑦 / 𝑧]𝜑))

This proof depends on syntax axioms:   → wi 4  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100

This theorem is used by: (None)