Theorem sblbis 1948

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 1947	. 2 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2		sblbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi2i 226	. 2 ⊢ (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))
4	1, 3	bitri 183	1 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))

Syntax hints:   ↔ wb 104  [wsb 1750

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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523

This theorem depends on definitions:  df-bi 116  df-nf 1449  df-sb 1751

This theorem is referenced by:  sb8eu  2027  sb8euh  2037  sb8iota  5160