Theorem sbrbis 1935

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)

Hypothesis

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

Assertion

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

Proof of Theorem sbrbis

Step	Hyp	Ref	Expression
1		sbbi 1933	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2		sbrbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi1i 227	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
4	1, 3	bitri 183	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   ↔ wb 104  [wsb 1736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737

This theorem is referenced by:  sbrbif  1936  sbabel  2308