Theorem ax11inda2 1368

Description: Induction step for constructing a substitution instance of ax-11o 1216 without using ax-11o 1216. Quantification case. When z and y are distinct, this theorem avoids the dummy variables needed by the more general ax11inda 1369.

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

Proof of Theorem ax11inda2

Step	Hyp	Ref	Expression
1		a16g 1274	. . . . 5 ⊢ (∀y y = z → ((x = y → ∀zφ) → ∀x(x = y → ∀zφ)))
2		ax-1 4	. . . . 5 ⊢ (∀zφ → (x = y → ∀zφ))
3	1, 2	syl5 21	. . . 4 ⊢ (∀y y = z → (∀zφ → ∀x(x = y → ∀zφ)))
4	3	a1d 12	. . 3 ⊢ (∀y y = z → (x = y → (∀zφ → ∀x(x = y → ∀zφ))))
5	4	a1d 12	. 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 1366	. 2 ⊢ (¬ ∀y y = z → (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ)))))
8	5, 7	pm2.61i 126	1 ⊢ (¬ ∀x x = y → (x = y → (∀zφ → ∀x(x = y → ∀zφ))))

Syntax hints:  ¬ wn 2   → wi 3  ∀wal 952   = wceq 954

This theorem is referenced by:  ax11inda 1369

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-9 963  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208

This theorem depends on definitions:  df-bi 147  df-an 225

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