Theorem ax11indi 2196

Description: Induction step for constructing a substitution instance of ax-11o 2141 without using ax-11o 2141. Implication case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
ax11indn.1	⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
ax11indi.2	⊢ (¬ ∀x x = y → (x = y → (ψ → ∀x(x = y → ψ))))

Assertion

Ref	Expression
ax11indi	⊢ (¬ ∀x x = y → (x = y → ((φ → ψ) → ∀x(x = y → (φ → ψ)))))

Proof of Theorem ax11indi

Step	Hyp	Ref	Expression
1		ax11indn.1	. . . . . 6 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
2	1	ax11indn 2195	. . . . 5 ⊢ (¬ ∀x x = y → (x = y → (¬ φ → ∀x(x = y → ¬ φ))))
3	2	imp 418	. . . 4 ⊢ ((¬ ∀x x = y ∧ x = y) → (¬ φ → ∀x(x = y → ¬ φ)))
4		pm2.21 100	. . . . . 6 ⊢ (¬ φ → (φ → ψ))
5	4	imim2i 13	. . . . 5 ⊢ ((x = y → ¬ φ) → (x = y → (φ → ψ)))
6	5	alimi 1559	. . . 4 ⊢ (∀x(x = y → ¬ φ) → ∀x(x = y → (φ → ψ)))
7	3, 6	syl6 29	. . 3 ⊢ ((¬ ∀x x = y ∧ x = y) → (¬ φ → ∀x(x = y → (φ → ψ))))
8		ax11indi.2	. . . . 5 ⊢ (¬ ∀x x = y → (x = y → (ψ → ∀x(x = y → ψ))))
9	8	imp 418	. . . 4 ⊢ ((¬ ∀x x = y ∧ x = y) → (ψ → ∀x(x = y → ψ)))
10		ax-1 6	. . . . . 6 ⊢ (ψ → (φ → ψ))
11	10	imim2i 13	. . . . 5 ⊢ ((x = y → ψ) → (x = y → (φ → ψ)))
12	11	alimi 1559	. . . 4 ⊢ (∀x(x = y → ψ) → ∀x(x = y → (φ → ψ)))
13	9, 12	syl6 29	. . 3 ⊢ ((¬ ∀x x = y ∧ x = y) → (ψ → ∀x(x = y → (φ → ψ))))
14	7, 13	jad 154	. 2 ⊢ ((¬ ∀x x = y ∧ x = y) → ((φ → ψ) → ∀x(x = y → (φ → ψ))))
15	14	ex 423	1 ⊢ (¬ ∀x x = y → (x = y → ((φ → ψ) → ∀x(x = y → (φ → ψ)))))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 358  ∀wal 1540

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542

This theorem is referenced by: (None)