Theorem sbrbis 1961

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 1959	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2		sbrbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi1i 228	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
4	1, 3	bitri 184	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   ↔ wb 105  [wsb 1762

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763

This theorem is referenced by:  sbrbif  1962  sbabel  2346