Theorem equs5e 1888

Description: A property related to substitution that unlike equs5 1996 doesn't require a distinctor antecedent. (Contributed by NM, 2-Feb-2007.)

Assertion

Ref	Expression
equs5e	⊢ (∃x(x = y ∧ φ) → ∀x(x = y → ∃yφ))

Proof of Theorem equs5e

Step	Hyp	Ref	Expression
1		nfe1 1732	. 2 ⊢ Ⅎx∃x(x = y ∧ φ)
2		equs3 1644	. . 3 ⊢ (∃x(x = y ∧ φ) ↔ ¬ ∀x(x = y → ¬ φ))
3		ax-11 1746	. . . . 5 ⊢ (x = y → (∀y ¬ φ → ∀x(x = y → ¬ φ)))
4	3	con3rr3 128	. . . 4 ⊢ (¬ ∀x(x = y → ¬ φ) → (x = y → ¬ ∀y ¬ φ))
5		df-ex 1542	. . . 4 ⊢ (∃yφ ↔ ¬ ∀y ¬ φ)
6	4, 5	syl6ibr 218	. . 3 ⊢ (¬ ∀x(x = y → ¬ φ) → (x = y → ∃yφ))
7	2, 6	sylbi 187	. 2 ⊢ (∃x(x = y ∧ φ) → (x = y → ∃yφ))
8	1, 7	alrimi 1765	1 ⊢ (∃x(x = y ∧ φ) → ∀x(x = y → ∃yφ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358  ∀wal 1540  ∃wex 1541

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This proof depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-nf 1545

This theorem is used by:  sb4e  1924