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 ( ( 𝐴S𝐵S ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )

Proof

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