Metamath Proof Explorer


Theorem norecdiv

Description: If a surreal has a reciprocal, then it has any division. (Contributed by Scott Fenton, 12-Mar-2025)

Ref Expression
Assertion norecdiv ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ∃ 𝑥 No ( 𝐴 ·s 𝑥 ) = 1s ) → ∃ 𝑦 No ( 𝐴 ·s 𝑦 ) = 𝐵 )

Proof

Step Hyp Ref Expression
1 simprl ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → 𝑤 No )
2 simpl3 ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → 𝐵 No )
3 1 2 mulscld ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ( 𝑤 ·s 𝐵 ) ∈ No )
4 oveq1 ( ( 𝐴 ·s 𝑤 ) = 1s → ( ( 𝐴 ·s 𝑤 ) ·s 𝐵 ) = ( 1s ·s 𝐵 ) )
5 4 adantl ( ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) → ( ( 𝐴 ·s 𝑤 ) ·s 𝐵 ) = ( 1s ·s 𝐵 ) )
6 5 adantl ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ( ( 𝐴 ·s 𝑤 ) ·s 𝐵 ) = ( 1s ·s 𝐵 ) )
7 simpl1 ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → 𝐴 No )
8 7 1 2 mulsassd ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ( ( 𝐴 ·s 𝑤 ) ·s 𝐵 ) = ( 𝐴 ·s ( 𝑤 ·s 𝐵 ) ) )
9 2 mulslidd ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ( 1s ·s 𝐵 ) = 𝐵 )
10 6 8 9 3eqtr3d ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ( 𝐴 ·s ( 𝑤 ·s 𝐵 ) ) = 𝐵 )
11 oveq2 ( 𝑧 = ( 𝑤 ·s 𝐵 ) → ( 𝐴 ·s 𝑧 ) = ( 𝐴 ·s ( 𝑤 ·s 𝐵 ) ) )
12 11 eqeq1d ( 𝑧 = ( 𝑤 ·s 𝐵 ) → ( ( 𝐴 ·s 𝑧 ) = 𝐵 ↔ ( 𝐴 ·s ( 𝑤 ·s 𝐵 ) ) = 𝐵 ) )
13 12 rspcev ( ( ( 𝑤 ·s 𝐵 ) ∈ No ∧ ( 𝐴 ·s ( 𝑤 ·s 𝐵 ) ) = 𝐵 ) → ∃ 𝑧 No ( 𝐴 ·s 𝑧 ) = 𝐵 )
14 3 10 13 syl2anc ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ( 𝑤 No ∧ ( 𝐴 ·s 𝑤 ) = 1s ) ) → ∃ 𝑧 No ( 𝐴 ·s 𝑧 ) = 𝐵 )
15 14 rexlimdvaa ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) → ( ∃ 𝑤 No ( 𝐴 ·s 𝑤 ) = 1s → ∃ 𝑧 No ( 𝐴 ·s 𝑧 ) = 𝐵 ) )
16 15 imp ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ∃ 𝑤 No ( 𝐴 ·s 𝑤 ) = 1s ) → ∃ 𝑧 No ( 𝐴 ·s 𝑧 ) = 𝐵 )
17 oveq2 ( 𝑥 = 𝑤 → ( 𝐴 ·s 𝑥 ) = ( 𝐴 ·s 𝑤 ) )
18 17 eqeq1d ( 𝑥 = 𝑤 → ( ( 𝐴 ·s 𝑥 ) = 1s ↔ ( 𝐴 ·s 𝑤 ) = 1s ) )
19 18 cbvrexvw ( ∃ 𝑥 No ( 𝐴 ·s 𝑥 ) = 1s ↔ ∃ 𝑤 No ( 𝐴 ·s 𝑤 ) = 1s )
20 19 anbi2i ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ∃ 𝑥 No ( 𝐴 ·s 𝑥 ) = 1s ) ↔ ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ∃ 𝑤 No ( 𝐴 ·s 𝑤 ) = 1s ) )
21 oveq2 ( 𝑦 = 𝑧 → ( 𝐴 ·s 𝑦 ) = ( 𝐴 ·s 𝑧 ) )
22 21 eqeq1d ( 𝑦 = 𝑧 → ( ( 𝐴 ·s 𝑦 ) = 𝐵 ↔ ( 𝐴 ·s 𝑧 ) = 𝐵 ) )
23 22 cbvrexvw ( ∃ 𝑦 No ( 𝐴 ·s 𝑦 ) = 𝐵 ↔ ∃ 𝑧 No ( 𝐴 ·s 𝑧 ) = 𝐵 )
24 16 20 23 3imtr4i ( ( ( 𝐴 No 𝐴 ≠ 0s𝐵 No ) ∧ ∃ 𝑥 No ( 𝐴 ·s 𝑥 ) = 1s ) → ∃ 𝑦 No ( 𝐴 ·s 𝑦 ) = 𝐵 )