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Metamath Proof Explorer

Theorem ssmd1

Description: Ordering implies the modular pair property. Remark in MaedaMaeda p. 1. (Contributed by NM, 21-Jun-2004) (New usage is discouraged.)

Ref Expression
Assertion ssmd1 ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐴 ⊆ 𝐵 ) → 𝐴 𝑀ℋ 𝐵 )

Proof

Step Hyp Ref Expression
1 inss1 ⊢ ( ( 𝑥 ∨ℋ 𝐴 ) ∩ 𝐵 ) ⊆ ( 𝑥 ∨ℋ 𝐴 )
2 dfss ⊢ ( 𝐴 ⊆ 𝐵 ↔ 𝐴 = ( 𝐴 ∩ 𝐵 ) )
3 2 biimpi ⊢ ( 𝐴 ⊆ 𝐵 → 𝐴 = ( 𝐴 ∩ 𝐵 ) )
4 3 oveq2d ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∨ℋ 𝐴 ) = ( 𝑥 ∨ℋ ( 𝐴 ∩ 𝐵 ) ) )
5 1 4 sseqtrid ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ∨ℋ 𝐴 ) ∩ 𝐵 ) ⊆ ( 𝑥 ∨ℋ ( 𝐴 ∩ 𝐵 ) ) )
6 5 a1d ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ⊆ 𝐵 → ( ( 𝑥 ∨ℋ 𝐴 ) ∩ 𝐵 ) ⊆ ( 𝑥 ∨ℋ ( 𝐴 ∩ 𝐵 ) ) ) )
7 6 ralrimivw ⊢ ( 𝐴 ⊆ 𝐵 → ∀ 𝑥 ∈ Cℋ ( 𝑥 ⊆ 𝐵 → ( ( 𝑥 ∨ℋ 𝐴 ) ∩ 𝐵 ) ⊆ ( 𝑥 ∨ℋ ( 𝐴 ∩ 𝐵 ) ) ) )
8 mdbr2 ⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ) → ( 𝐴 𝑀ℋ 𝐵 ↔ ∀ 𝑥 ∈ Cℋ ( 𝑥 ⊆ 𝐵 → ( ( 𝑥 ∨ℋ 𝐴 ) ∩ 𝐵 ) ⊆ ( 𝑥 ∨ℋ ( 𝐴 ∩ 𝐵 ) ) ) ) )
9 7 8 imbitrrid ⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ) → ( 𝐴 ⊆ 𝐵 → 𝐴 𝑀ℋ 𝐵 ) )
10 9 3impia ⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐴 ⊆ 𝐵 ) → 𝐴 𝑀ℋ 𝐵 )