Metamath Proof Explorer


Theorem nfrald

Description: Deduction version of nfral . Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker nfraldw when possible. (Contributed by NM, 15-Feb-2013) (Revised by Mario Carneiro, 7-Oct-2016) (New usage is discouraged.)

Ref Expression
Hypotheses nfrald.1 ⊢ Ⅎ 𝑦 𝜑
nfrald.2 ⊢ ( 𝜑 → Ⅎ 𝑥 𝐴 )
nfrald.3 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
Assertion nfrald ( 𝜑 → Ⅎ 𝑥 ∀ 𝑦 ∈ 𝐴 𝜓 )

Proof

Step Hyp Ref Expression
1 nfrald.1 ⊢ Ⅎ 𝑦 𝜑
2 nfrald.2 ⊢ ( 𝜑 → Ⅎ 𝑥 𝐴 )
3 nfrald.3 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
4 df-ral ⊢ ( ∀ 𝑦 ∈ 𝐴 𝜓 ↔ ∀ 𝑦 ( 𝑦 ∈ 𝐴 → 𝜓 ) )
5 nfcvf ⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → Ⅎ 𝑥 𝑦 )
6 5 adantl ⊢ ( ( 𝜑 ∧ ¬ ∀ 𝑥 𝑥 = 𝑦 ) → Ⅎ 𝑥 𝑦 )
7 2 adantr ⊢ ( ( 𝜑 ∧ ¬ ∀ 𝑥 𝑥 = 𝑦 ) → Ⅎ 𝑥 𝐴 )
8 6 7 nfeld ⊢ ( ( 𝜑 ∧ ¬ ∀ 𝑥 𝑥 = 𝑦 ) → Ⅎ 𝑥 𝑦 ∈ 𝐴 )
9 3 adantr ⊢ ( ( 𝜑 ∧ ¬ ∀ 𝑥 𝑥 = 𝑦 ) → Ⅎ 𝑥 𝜓 )
10 8 9 nfimd ⊢ ( ( 𝜑 ∧ ¬ ∀ 𝑥 𝑥 = 𝑦 ) → Ⅎ 𝑥 ( 𝑦 ∈ 𝐴 → 𝜓 ) )
11 1 10 nfald2 ⊢ ( 𝜑 → Ⅎ 𝑥 ∀ 𝑦 ( 𝑦 ∈ 𝐴 → 𝜓 ) )
12 4 11 nfxfrd ⊢ ( 𝜑 → Ⅎ 𝑥 ∀ 𝑦 ∈ 𝐴 𝜓 )