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Metamath Proof Explorer

Theorem shscom

Description: Commutative law for subspace sum. (Contributed by NM, 15-Dec-2004) (New usage is discouraged.)

Ref Expression
Assertion shscom ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐴 +ℋ 𝐵 ) = ( 𝐵 +ℋ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 shel ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝑦 ∈ 𝐴 ) → 𝑦 ∈ ℋ )
2 shel ⊢ ( ( 𝐵 ∈ Sℋ ∧ 𝑧 ∈ 𝐵 ) → 𝑧 ∈ ℋ )
3 1 2 anim12i ⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝑦 ∈ 𝐴 ) ∧ ( 𝐵 ∈ Sℋ ∧ 𝑧 ∈ 𝐵 ) ) → ( 𝑦 ∈ ℋ ∧ 𝑧 ∈ ℋ ) )
4 3 an4s ⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵 ) ) → ( 𝑦 ∈ ℋ ∧ 𝑧 ∈ ℋ ) )
5 ax-hvcom ⊢ ( ( 𝑦 ∈ ℋ ∧ 𝑧 ∈ ℋ ) → ( 𝑦 +ℎ 𝑧 ) = ( 𝑧 +ℎ 𝑦 ) )
6 4 5 syl ⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵 ) ) → ( 𝑦 +ℎ 𝑧 ) = ( 𝑧 +ℎ 𝑦 ) )
7 6 eqeq2d ⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵 ) ) → ( 𝑥 = ( 𝑦 +ℎ 𝑧 ) ↔ 𝑥 = ( 𝑧 +ℎ 𝑦 ) ) )
8 7 2rexbidva ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 𝑥 = ( 𝑦 +ℎ 𝑧 ) ↔ ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 𝑥 = ( 𝑧 +ℎ 𝑦 ) ) )
9 rexcom ⊢ ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 𝑥 = ( 𝑧 +ℎ 𝑦 ) ↔ ∃ 𝑧 ∈ 𝐵 ∃ 𝑦 ∈ 𝐴 𝑥 = ( 𝑧 +ℎ 𝑦 ) )
10 8 9 bitrdi ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 𝑥 = ( 𝑦 +ℎ 𝑧 ) ↔ ∃ 𝑧 ∈ 𝐵 ∃ 𝑦 ∈ 𝐴 𝑥 = ( 𝑧 +ℎ 𝑦 ) ) )
11 shsel ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐴 +ℋ 𝐵 ) ↔ ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
12 shsel ⊢ ( ( 𝐵 ∈ Sℋ ∧ 𝐴 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐵 +ℋ 𝐴 ) ↔ ∃ 𝑧 ∈ 𝐵 ∃ 𝑦 ∈ 𝐴 𝑥 = ( 𝑧 +ℎ 𝑦 ) ) )
13 12 ancoms ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐵 +ℋ 𝐴 ) ↔ ∃ 𝑧 ∈ 𝐵 ∃ 𝑦 ∈ 𝐴 𝑥 = ( 𝑧 +ℎ 𝑦 ) ) )
14 10 11 13 3bitr4d ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐴 +ℋ 𝐵 ) ↔ 𝑥 ∈ ( 𝐵 +ℋ 𝐴 ) ) )
15 14 eqrdv ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐴 +ℋ 𝐵 ) = ( 𝐵 +ℋ 𝐴 ) )