Theorem ax11indi 1365

Description: Induction step for constructing a substitution instance of ax-11o 1216 without using ax-11o 1216. Implication case.

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	. . . . . . 7 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
2	1	ax11indn 1364	. . . . . 6 ⊢ (¬ ∀x x = y → (x = y → (¬ φ → ∀x(x = y → ¬ φ))))
3	2	imp 350	. . . . 5 ⊢ ((¬ ∀x x = y ⋀ x = y) → (¬ φ → ∀x(x = y → ¬ φ)))
4		pm2.21 76	. . . . . . 7 ⊢ (¬ φ → (φ → ψ))
5	4	imim2i 17	. . . . . 6 ⊢ ((x = y → ¬ φ) → (x = y → (φ → ψ)))
6	5	19.20i 990	. . . . 5 ⊢ (∀x(x = y → ¬ φ) → ∀x(x = y → (φ → ψ)))
7	3, 6	syl6 22	. . . 4 ⊢ ((¬ ∀x x = y ⋀ x = y) → (¬ φ → ∀x(x = y → (φ → ψ))))
8		ax11indi.2	. . . . . 6 ⊢ (¬ ∀x x = y → (x = y → (ψ → ∀x(x = y → ψ))))
9	8	imp 350	. . . . 5 ⊢ ((¬ ∀x x = y ⋀ x = y) → (ψ → ∀x(x = y → ψ)))
10		ax-1 4	. . . . . . 7 ⊢ (ψ → (φ → ψ))
11	10	imim2i 17	. . . . . 6 ⊢ ((x = y → ψ) → (x = y → (φ → ψ)))
12	11	19.20i 990	. . . . 5 ⊢ (∀x(x = y → ψ) → ∀x(x = y → (φ → ψ)))
13	9, 12	syl6 22	. . . 4 ⊢ ((¬ ∀x x = y ⋀ x = y) → (ψ → ∀x(x = y → (φ → ψ))))
14	7, 13	jaod 424	. . 3 ⊢ ((¬ ∀x x = y ⋀ x = y) → ((¬ φ ⋁ ψ) → ∀x(x = y → (φ → ψ))))
15		imor 234	. . 3 ⊢ ((φ → ψ) ↔ (¬ φ ⋁ ψ))
16	14, 15	syl5ib 206	. 2 ⊢ ((¬ ∀x x = y ⋀ x = y) → ((φ → ψ) → ∀x(x = y → (φ → ψ))))
17	16	ex 373	1 ⊢ (¬ ∀x x = y → (x = y → ((φ → ψ) → ∀x(x = y → (φ → ψ)))))

Syntax hints:  ¬ wn 2   → wi 3   ⋁ wo 222   ⋀ wa 223  ∀wal 952   = wceq 954

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 961  ax-4 971  ax-5o 973  ax-6o 976

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 979

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