Theorem bnj1109 32794

Description: First-order logic and set theory. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)

Hypotheses

Ref	Expression
bnj1109.1	⊢ ∃𝑥((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓)
bnj1109.2	⊢ ((𝐴 = 𝐵 ∧ 𝜑) → 𝜓)

Assertion

Ref	Expression
bnj1109	⊢ ∃𝑥(𝜑 → 𝜓)

Proof of Theorem bnj1109

Step	Hyp	Ref	Expression
1		bnj1109.2	. . . . . . 7 ⊢ ((𝐴 = 𝐵 ∧ 𝜑) → 𝜓)
2	1	ex 412	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝜑 → 𝜓))
3	2	a1i 11	. . . . 5 ⊢ ((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → (𝐴 = 𝐵 → (𝜑 → 𝜓)))
4	3	ax-gen 1793	. . . 4 ⊢ ∀𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → (𝐴 = 𝐵 → (𝜑 → 𝜓)))
5		bnj1109.1	. . . . 5 ⊢ ∃𝑥((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓)
6		impexp 450	. . . . . 6 ⊢ (((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓) ↔ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
7	6	exbii 1846	. . . . 5 ⊢ (∃𝑥((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓) ↔ ∃𝑥(𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
8	5, 7	mpbi 229	. . . 4 ⊢ ∃𝑥(𝐴 ≠ 𝐵 → (𝜑 → 𝜓))
9		exintr 1891	. . . 4 ⊢ (∀𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → (𝐴 = 𝐵 → (𝜑 → 𝜓))) → (∃𝑥(𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → ∃𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 = 𝐵 → (𝜑 → 𝜓)))))
10	4, 8, 9	mp2 9	. . 3 ⊢ ∃𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 = 𝐵 → (𝜑 → 𝜓)))
11		exancom 1860	. . 3 ⊢ (∃𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 = 𝐵 → (𝜑 → 𝜓))) ↔ ∃𝑥((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓))))
12	10, 11	mpbi 229	. 2 ⊢ ∃𝑥((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
13		df-ne 2939	. . . 4 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
14	13	imbi1i 349	. . 3 ⊢ ((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ↔ (¬ 𝐴 = 𝐵 → (𝜑 → 𝜓)))
15		pm2.61 191	. . . 4 ⊢ ((𝐴 = 𝐵 → (𝜑 → 𝜓)) → ((¬ 𝐴 = 𝐵 → (𝜑 → 𝜓)) → (𝜑 → 𝜓)))
16	15	imp 406	. . 3 ⊢ (((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (¬ 𝐴 = 𝐵 → (𝜑 → 𝜓))) → (𝜑 → 𝜓))
17	14, 16	sylan2b 593	. 2 ⊢ (((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓))) → (𝜑 → 𝜓))
18	12, 17	bnj101 32730	1 ⊢ ∃𝑥(𝜑 → 𝜓)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  ∀wal 1535   = wceq 1537  ∃wex 1777   ≠ wne 2938

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1778  df-ne 2939

This theorem is referenced by:  bnj1030  32995  bnj1128  32998  bnj1145  33001