Theorem nfrandd-P6 816

Description: ENF Over Conjunction (deductive form).

Hypotheses

Ref	Expression
nfrandd-P6.1	⊢ (𝛾 → Ⅎ𝑥𝜑)
nfrandd-P6.2	⊢ (𝛾 → Ⅎ𝑥𝜓)

Assertion

Ref	Expression
nfrandd-P6	⊢ (𝛾 → Ⅎ𝑥(𝜑 ∧ 𝜓))

Proof of Theorem nfrandd-P6

Step	Hyp	Ref	Expression
1		nfrandd-P6.1	. . . 4 ⊢ (𝛾 → Ⅎ𝑥𝜑)
2		nfrandd-P6.2	. . . . 5 ⊢ (𝛾 → Ⅎ𝑥𝜓)
3		nfrneg-P6 688	. . . . . 6 ⊢ (Ⅎ𝑥 ¬ 𝜓 ↔ Ⅎ𝑥𝜓)
4	3	bisym-P3.33b.RC 299	. . . . 5 ⊢ (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥 ¬ 𝜓)
5	2, 4	subimr2-P4.RC 543	. . . 4 ⊢ (𝛾 → Ⅎ𝑥 ¬ 𝜓)
6	1, 5	nfrimd-P6 815	. . 3 ⊢ (𝛾 → Ⅎ𝑥(𝜑 → ¬ 𝜓))
7		nfrneg-P6 688	. . . 4 ⊢ (Ⅎ𝑥 ¬ (𝜑 → ¬ 𝜓) ↔ Ⅎ𝑥(𝜑 → ¬ 𝜓))
8	7	bisym-P3.33b.RC 299	. . 3 ⊢ (Ⅎ𝑥(𝜑 → ¬ 𝜓) ↔ Ⅎ𝑥 ¬ (𝜑 → ¬ 𝜓))
9	6, 8	subimr2-P4.RC 543	. 2 ⊢ (𝛾 → Ⅎ𝑥 ¬ (𝜑 → ¬ 𝜓))
10		andasim-P3.46a 356	. . . 4 ⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 → ¬ 𝜓))
11	10	nfrleq-P6 687	. . 3 ⊢ (Ⅎ𝑥(𝜑 ∧ 𝜓) ↔ Ⅎ𝑥 ¬ (𝜑 → ¬ 𝜓))
12	11	bisym-P3.33b.RC 299	. 2 ⊢ (Ⅎ𝑥 ¬ (𝜑 → ¬ 𝜓) ↔ Ⅎ𝑥(𝜑 ∧ 𝜓))
13	9, 12	subimr2-P4.RC 543	1 ⊢ (𝛾 → Ⅎ𝑥(𝜑 ∧ 𝜓))

Syntax hints:  ¬ wff-neg 9   → wff-imp 10   ∧ wff-and 132  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:  nfrbid-P6  818  ndnfrand-P7.4  829