Metamath Proof Explorer


Theorem selvply1rhmlem5

Description: Lemma for selvply1rhm . (Contributed by Thierry Arnoux, 4-May-2026)

Ref Expression
Hypotheses selvply1rhm.1 B = Base P
selvply1rhm.2 P = I mPoly R
selvply1rhm.3 U = I X mPoly R
selvply1rhm.4 Q = Poly 1 U
selvply1rhm.5 H = f B n 0 1 𝑜 I selectVars R X f X n
selvply1rhm.6 φ I V
selvply1rhm.7 φ X I
selvply1rhm.8 φ R CRing
selvply1rhmlem5.f φ F B
selvply1rhmlem5.m M = q Base X mPoly U s 0 1 𝑜 q X s
Assertion selvply1rhmlem5 φ H F = M I selectVars R X F

Proof

Step Hyp Ref Expression
1 selvply1rhm.1 B = Base P
2 selvply1rhm.2 P = I mPoly R
3 selvply1rhm.3 U = I X mPoly R
4 selvply1rhm.4 Q = Poly 1 U
5 selvply1rhm.5 H = f B n 0 1 𝑜 I selectVars R X f X n
6 selvply1rhm.6 φ I V
7 selvply1rhm.7 φ X I
8 selvply1rhm.8 φ R CRing
9 selvply1rhmlem5.f φ F B
10 selvply1rhmlem5.m M = q Base X mPoly U s 0 1 𝑜 q X s
11 fveq2 f = F I selectVars R X f = I selectVars R X F
12 11 fveq1d f = F I selectVars R X f X n = I selectVars R X F X n
13 12 mpteq2dv f = F n 0 1 𝑜 I selectVars R X f X n = n 0 1 𝑜 I selectVars R X F X n
14 ovexd φ 0 1 𝑜 V
15 14 mptexd φ n 0 1 𝑜 I selectVars R X F X n V
16 5 13 9 15 fvmptd3 φ H F = n 0 1 𝑜 I selectVars R X F X n
17 fveq1 s = n s = n
18 17 opeq2d s = n X s = X n
19 18 sneqd s = n X s = X n
20 19 fveq2d s = n q X s = q X n
21 20 cbvmptv s 0 1 𝑜 q X s = n 0 1 𝑜 q X n
22 21 mpteq2i q Base X mPoly U s 0 1 𝑜 q X s = q Base X mPoly U n 0 1 𝑜 q X n
23 10 22 eqtri M = q Base X mPoly U n 0 1 𝑜 q X n
24 fveq1 q = I selectVars R X F q X n = I selectVars R X F X n
25 24 mpteq2dv q = I selectVars R X F n 0 1 𝑜 q X n = n 0 1 𝑜 I selectVars R X F X n
26 eqid X mPoly U = X mPoly U
27 eqid Base X mPoly U = Base X mPoly U
28 7 snssd φ X I
29 2 1 3 26 27 8 28 9 selvcl φ I selectVars R X F Base X mPoly U
30 23 25 29 15 fvmptd3 φ M I selectVars R X F = n 0 1 𝑜 I selectVars R X F X n
31 16 30 eqtr4d φ H F = M I selectVars R X F