Theorem ax11f 2192

Description: Basis step for constructing a substitution instance of ax-11o 2141 without using ax-11o 2141. We can start with any formula φ in which x is not free. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
ax11f.1	⊢ (φ → ∀xφ)

Assertion

Ref	Expression
ax11f	⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))

Proof of Theorem ax11f

Step	Hyp	Ref	Expression
1		ax11f.1	. . 3 ⊢ (φ → ∀xφ)
2		ax-1 6	. . 3 ⊢ (φ → (x = y → φ))
3	1, 2	alrimih 1565	. 2 ⊢ (φ → ∀x(x = y → φ))
4	3	2a1i 24	1 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4  ∀wal 1540

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-gen 1546  ax-5 1557

This theorem is used by: (None)