Metamath Proof Explorer


Theorem lsmsnpridl

Description: The product of the ring with a single element is equal to the principal ideal generated by that element. (Contributed by Thierry Arnoux, 21-Jan-2024)

Ref Expression
Hypotheses lsmsnpridl.1 B = Base R
lsmsnpridl.2 G = mulGrp R
lsmsnpridl.3 × ˙ = LSSum G
lsmsnpridl.4 K = RSpan R
lsmsnpridl.5 φ R Ring
lsmsnpridl.6 φ X B
Assertion lsmsnpridl φ B × ˙ X = K X

Proof

Step Hyp Ref Expression
1 lsmsnpridl.1 B = Base R
2 lsmsnpridl.2 G = mulGrp R
3 lsmsnpridl.3 × ˙ = LSSum G
4 lsmsnpridl.4 K = RSpan R
5 lsmsnpridl.5 φ R Ring
6 lsmsnpridl.6 φ X B
7 2 1 mgpbas B = Base G
8 eqid R = R
9 2 8 mgpplusg R = + G
10 2 fvexi G V
11 10 a1i φ G V
12 ssidd φ B B
13 7 9 3 11 12 6 elgrplsmsn φ x B × ˙ X y B x = y R X
14 1 8 4 rspsnel R Ring X B x K X y B x = y R X
15 5 6 14 syl2anc φ x K X y B x = y R X
16 13 15 bitr4d φ x B × ˙ X x K X
17 16 eqrdv φ B × ˙ X = K X