Theorem bnj1109 35417

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 418	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝜑 → 𝜓))
3	2	a1i 11	. . . . 5 ⊢ ((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → (𝐴 = 𝐵 → (𝜑 → 𝜓)))
4	3	ax-gen 1827	. . . 4 ⊢ ∀𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) → (𝐴 = 𝐵 → (𝜑 → 𝜓)))
5		bnj1109.1	. . . . 5 ⊢ ∃𝑥((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓)
6		impexp 456	. . . . . 6 ⊢ (((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓) ↔ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
7	6	exbii 1880	. . . . 5 ⊢ (∃𝑥((𝐴 ≠ 𝐵 ∧ 𝜑) → 𝜓) ↔ ∃𝑥(𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
8	5, 7	mpbi 233	. . . 4 ⊢ ∃𝑥(𝐴 ≠ 𝐵 → (𝜑 → 𝜓))
9		exintr 1924	. . . 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 1893	. . 3 ⊢ (∃𝑥((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 = 𝐵 → (𝜑 → 𝜓))) ↔ ∃𝑥((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓))))
12	10, 11	mpbi 233	. 2 ⊢ ∃𝑥((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓)))
13		df-ne 2956	. . . 4 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
14	13	imbi1i 352	. . 3 ⊢ ((𝐴 ≠ 𝐵 → (𝜑 → 𝜓)) ↔ (¬ 𝐴 = 𝐵 → (𝜑 → 𝜓)))
15		pm2.61 194	. . . 4 ⊢ ((𝐴 = 𝐵 → (𝜑 → 𝜓)) → ((¬ 𝐴 = 𝐵 → (𝜑 → 𝜓)) → (𝜑 → 𝜓)))
16	15	imp 412	. . 3 ⊢ (((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (¬ 𝐴 = 𝐵 → (𝜑 → 𝜓))) → (𝜑 → 𝜓))
17	14, 16	sylan2b 606	. 2 ⊢ (((𝐴 = 𝐵 → (𝜑 → 𝜓)) ∧ (𝐴 ≠ 𝐵 → (𝜑 → 𝜓))) → (𝜑 → 𝜓))
18	12, 17	bnj101 35354	1 ⊢ ∃𝑥(𝜑 → 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1567   = wceq 1569  ∃wex 1811   ≠ wne 2955

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-ne 2956

This theorem is used by:  bnj1030  35617  bnj1128  35620  bnj1145  35623