Theorem sblbis 1972

Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993.)

Hypothesis

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

Assertion

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

Proof of Theorem sblbis

Step	Hyp	Ref	Expression
1		sbbi 1971	. 2 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2		sblbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi2i 227	. 2 ⊢ (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))
4	1, 3	bitri 184	1 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))

Syntax hints:   ↔ wb 105  [wsb 1773

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774

This theorem is referenced by:  sb8eu  2051  sb8euh  2061  sb8iota  5203