Theorem sblbisvOLD 2329

Description: Obsolete version of sblbis 2319 as of 24-Jul-2023. Introduce left biconditional inside of a substitution. Version of sblbis 2319 with a disjoint variable condition, not requiring ax-13 2390. (Contributed by Wolf Lammen, 18-Jan-2023.) (New usage is discouraged.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
sblbisvOLD.1	⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)

Assertion

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

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem sblbisvOLD

Step	Hyp	Ref	Expression
1		sbbivOLD 2327	. 2 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2		sblbisvOLD.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi2i 340	. 2 ⊢ (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))
4	1, 3	bitri 277	1 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))

Syntax hints:   ↔ wb 208  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785  df-sb 2070

This theorem is referenced by:  sbievOLD  2331