Theorem equs5aALT 2400

Description: Alternate proof of equs5a 2491. Uses ax-12 2216 but not ax-13 2406. (Contributed by NM, 2-Feb-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
equs5aALT	⊢ (∃𝑥(𝑥 = 𝑦 ∧ ∀𝑦𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))

Proof of Theorem equs5aALT

Step	Hyp	Ref	Expression
1		nfa1 2189	. 2 ⊢ Ⅎ𝑥∀𝑥(𝑥 = 𝑦 → 𝜑)
2		ax-12 2216	. . 3 ⊢ (𝑥 = 𝑦 → (∀𝑦𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3	2	imp 412	. 2 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑦𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))
4	1, 3	exlimi 2256	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ∀𝑦𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817

This theorem is used by: (None)