Metamath Proof Explorer


Theorem divsmulw

Description: Relationship between surreal division and multiplication. Weak version that does not assume reciprocals. Later, when we prove precsex , we can eliminate the existence hypothesis (see divsmul ). (Contributed by Scott Fenton, 12-Mar-2025)

Ref Expression
Assertion divsmulw A No B No C No C 0 s x No C s x = 1 s A / su C = B C s B = A

Proof

Step Hyp Ref Expression
1 divsval A No C No C 0 s A / su C = ι y No | C s y = A
2 1 eqeq1d A No C No C 0 s A / su C = B ι y No | C s y = A = B
3 2 3expb A No C No C 0 s A / su C = B ι y No | C s y = A = B
4 3 3adant2 A No B No C No C 0 s A / su C = B ι y No | C s y = A = B
5 4 adantr A No B No C No C 0 s x No C s x = 1 s A / su C = B ι y No | C s y = A = B
6 simpl2 A No B No C No C 0 s x No C s x = 1 s B No
7 simp3l A No B No C No C 0 s C No
8 simp3r A No B No C No C 0 s C 0 s
9 simp1 A No B No C No C 0 s A No
10 7 8 9 3jca A No B No C No C 0 s C No C 0 s A No
11 noreceuw C No C 0 s A No x No C s x = 1 s ∃! y No C s y = A
12 10 11 sylan A No B No C No C 0 s x No C s x = 1 s ∃! y No C s y = A
13 oveq2 y = B C s y = C s B
14 13 eqeq1d y = B C s y = A C s B = A
15 14 riota2 B No ∃! y No C s y = A C s B = A ι y No | C s y = A = B
16 6 12 15 syl2anc A No B No C No C 0 s x No C s x = 1 s C s B = A ι y No | C s y = A = B
17 5 16 bitr4d A No B No C No C 0 s x No C s x = 1 s A / su C = B C s B = A