Metamath Proof Explorer


Theorem bnj1109

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

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

Proof

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