Theorem sbbidvOLD 2084

Description: Obsolete version of sbbidv 2083 as of 6-Jul-2023. Deduction substituting both sides of a biconditional, with 𝜑 and 𝑥 disjoint. See also sbbid 2245. (Contributed by Wolf Lammen, 6-May-2023.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
sbbidv.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbbidvOLD	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑦)   𝜓(𝑥,𝑦)   𝜒(𝑥,𝑦)

Proof of Theorem sbbidvOLD

Step	Hyp	Ref	Expression
1		sbbidv.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	biimpd 231	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3	2	sbimdv 2082	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜒))
4	1	biimprd 250	. . 3 ⊢ (𝜑 → (𝜒 → 𝜓))
5	4	sbimdv 2082	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜒 → [𝑦 / 𝑥]𝜓))
6	3, 5	impbid 214	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 208  [wsb 2068

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

This theorem depends on definitions:  df-bi 209  df-sb 2069

This theorem is referenced by: (None)