Theorem lemma-L6.05a 764

Description: This lemma is the result of solvedsub-P6b 713 with no hypothesis required.

'𝑦' cannot occur in '𝑡'.

Hypothesis

Ref	Expression
lemma-L6.05a.1	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
lemma-L6.05a	⊢ (𝑥 = 𝑡 → (𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Distinct variable groups:   𝑡,𝑦   𝑥,𝑦

Proof of Theorem lemma-L6.05a

Step	Hyp	Ref	Expression
1		lemma-L6.05a.1	. 2 ⊢ Ⅎ𝑦𝜑
2		nfrall1-P6 741	. 2 ⊢ Ⅎ𝑦∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		lemma-L6.01a 724	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
4		lemma-L6.01a 724	. 2 ⊢ (𝑦 = 𝑡 → (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
5	1, 2, 3, 4	trnsvsub-P6 763	1 ⊢ (𝑥 = 𝑡 → (𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Syntax hints:  term-obj 1   = wff-equals 6  ∀wff-forall 8   → 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  ax-L5 17  ax-L6 18  ax-L7 19  ax-L10 27  ax-L12 29

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:  psubtoisub-P6  765