Metamath Proof Explorer


Theorem sgnsv

Description: The sign mapping. (Contributed by Thierry Arnoux, 9-Sep-2018)

Ref Expression
Hypotheses sgnsval.b ⊢ B = Base R
sgnsval.0 ⊢ 0 ˙ = 0 R
sgnsval.l ⊢ < ˙ = < R
sgnsval.s ⊢ S = sgn s ⁡ R
Assertion sgnsv ⊢ R ∈ V → S = x ∈ B ⟼ if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1

Proof

Step Hyp Ref Expression
1 sgnsval.b ⊢ B = Base R
2 sgnsval.0 ⊢ 0 ˙ = 0 R
3 sgnsval.l ⊢ < ˙ = < R
4 sgnsval.s ⊢ S = sgn s ⁡ R
5 elex ⊢ R ∈ V → R ∈ V
6 fveq2 ⊢ r = R → Base r = Base R
7 6 1 eqtr4di ⊢ r = R → Base r = B
8 fveq2 ⊢ r = R → 0 r = 0 R
9 8 2 eqtr4di ⊢ r = R → 0 r = 0 ˙
10 9 adantr ⊢ r = R ∧ x ∈ Base r → 0 r = 0 ˙
11 10 eqeq2d ⊢ r = R ∧ x ∈ Base r → x = 0 r ↔ x = 0 ˙
12 fveq2 ⊢ r = R → < r = < R
13 12 3 eqtr4di ⊢ r = R → < r = < ˙
14 13 adantr ⊢ r = R ∧ x ∈ Base r → < r = < ˙
15 eqidd ⊢ r = R ∧ x ∈ Base r → x = x
16 10 14 15 breq123d ⊢ r = R ∧ x ∈ Base r → 0 r < r x ↔ 0 ˙ < ˙ x
17 16 ifbid ⊢ r = R ∧ x ∈ Base r → if 0 r < r x 1 − 1 = if 0 ˙ < ˙ x 1 − 1
18 11 17 ifbieq2d ⊢ r = R ∧ x ∈ Base r → if x = 0 r 0 if 0 r < r x 1 − 1 = if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1
19 7 18 mpteq12dva ⊢ r = R → x ∈ Base r ⟼ if x = 0 r 0 if 0 r < r x 1 − 1 = x ∈ B ⟼ if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1
20 df-sgns ⊢ sgn s = r ∈ V ⟼ x ∈ Base r ⟼ if x = 0 r 0 if 0 r < r x 1 − 1
21 19 20 1 mptfvmpt ⊢ R ∈ V → sgn s ⁡ R = x ∈ B ⟼ if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1
22 5 21 syl ⊢ R ∈ V → sgn s ⁡ R = x ∈ B ⟼ if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1
23 4 22 eqtrid ⊢ R ∈ V → S = x ∈ B ⟼ if x = 0 ˙ 0 if 0 ˙ < ˙ x 1 − 1