Theorem ax11inda2 2199

Description: Induction step for constructing a substitution instance of ax-11o 2141 without using ax-11o 2141. Quantification case. When z and y are distinct, this theorem avoids the dummy variables needed by the more general ax11inda 2200. (Contributed by NM, 24-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
ax11inda2.1	⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))

Assertion

Ref	Expression
ax11inda2	⊢ (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ))))

Distinct variable group:   y,z

Allowed substitution hints:   φ(x, y, z)

Proof of Theorem ax11inda2

Step	Hyp	Ref	Expression
1		ax-1 6	. . . . 5 ⊢ (∀zφ → (x = y → ∀zφ))
2		a16g-o 2186	. . . . 5 ⊢ (∀y y = z → ((x = y → ∀zφ) → ∀x(x = y → ∀zφ)))
3	1, 2	syl5 28	. . . 4 ⊢ (∀y y = z → (∀zφ → ∀x(x = y → ∀zφ)))
4	3	a1d 22	. . 3 ⊢ (∀y y = z → (x = y → (∀zφ → ∀x(x = y → ∀zφ))))
5	4	a1d 22	. 2 ⊢ (∀y y = z → (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ)))))
6		ax11inda2.1	. . 3 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
7	6	ax11indalem 2197	. 2 ⊢ (¬ ∀y y = z → (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ)))))
8	5, 7	pm2.61i 156	1 ⊢ (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ))))

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-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-4 2135  ax-5o 2136  ax-6o 2137  ax-10o 2139  ax-12o 2142  ax-16 2144

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is used by:  ax11inda  2200