Theorem bj-sbidmOLD 36851

Description: Obsolete proof of sbidm 2515 temporarily kept here to check it gives no additional insight. (Contributed by NM, 8-Mar-1995.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
bj-sbidmOLD	⊢ ([𝑦 / 𝑥][𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)

Proof of Theorem bj-sbidmOLD

Step	Hyp	Ref	Expression
1		equsb2 2497	. . 3 ⊢ [𝑦 / 𝑥]𝑦 = 𝑥
2		sbequ12r 2252	. . . 4 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
3	2	sbimi 2074	. . 3 ⊢ ([𝑦 / 𝑥]𝑦 = 𝑥 → [𝑦 / 𝑥]([𝑦 / 𝑥]𝜑 ↔ 𝜑))
4	1, 3	ax-mp 5	. 2 ⊢ [𝑦 / 𝑥]([𝑦 / 𝑥]𝜑 ↔ 𝜑)
5		sbbi 2308	. 2 ⊢ ([𝑦 / 𝑥]([𝑦 / 𝑥]𝜑 ↔ 𝜑) ↔ ([𝑦 / 𝑥][𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑))
6	4, 5	mpbi 230	1 ⊢ ([𝑦 / 𝑥][𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)

Syntax hints:   ↔ wb 206  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-10 2141  ax-12 2177  ax-13 2377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-ex 1780  df-nf 1784  df-sb 2065

This theorem is referenced by: (None)