Theorem ndnfrbi-P7.6 831

Description: Natural Deduction: Effective Non-Freeness Rule 6.

If '𝑥' is (conditionally) effectively not free in both '𝜑' and '𝜓', then '𝑥' is (conditionally) effectively not free in '(𝜑 ↔ 𝜓)'.

Hypotheses

Ref	Expression
ndnfrbi-P7.6.1	⊢ (𝛾 → Ⅎ𝑥𝜑)
ndnfrbi-P7.6.2	⊢ (𝛾 → Ⅎ𝑥𝜓)

Assertion

Ref	Expression
ndnfrbi-P7.6	⊢ (𝛾 → Ⅎ𝑥(𝜑 ↔ 𝜓))

Proof of Theorem ndnfrbi-P7.6

Step	Hyp	Ref	Expression
1		ndnfrbi-P7.6.1	. 2 ⊢ (𝛾 → Ⅎ𝑥𝜑)
2		ndnfrbi-P7.6.2	. 2 ⊢ (𝛾 → Ⅎ𝑥𝜓)
3	1, 2	nfrbid-P6 818	1 ⊢ (𝛾 → Ⅎ𝑥(𝜑 ↔ 𝜓))

Syntax hints:   → wff-imp 10   ↔ wff-bi 104  Ⅎwff-nfree 681

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14  ax-GEN 15  ax-L4 16

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161  df-exists-D5.1 596  df-nfree-D6.1 682

This theorem is referenced by:  ndnfrbi-P7.6.RC  879  ndnfrbi-P7.6.CL  908  lemma-L7.03  962