Metamath Proof Explorer


Theorem swrdfv0

Description: The first symbol in an extracted subword. (Contributed by AV, 27-Apr-2022)

Ref Expression
Assertion swrdfv0 ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → S substr F L ⁡ 0 = S ⁡ F

Proof

Step Hyp Ref Expression
1 elfzofz ⊢ F ∈ 0 ..^ L → F ∈ 0 … L
2 1 3anim2i ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → S ∈ Word A ∧ F ∈ 0 … L ∧ L ∈ 0 … S
3 fzonnsub ⊢ F ∈ 0 ..^ L → L − F ∈ ℕ
4 3 3ad2ant2 ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → L − F ∈ ℕ
5 lbfzo0 ⊢ 0 ∈ 0 ..^ L − F ↔ L − F ∈ ℕ
6 4 5 sylibr ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → 0 ∈ 0 ..^ L − F
7 swrdfv ⊢ S ∈ Word A ∧ F ∈ 0 … L ∧ L ∈ 0 … S ∧ 0 ∈ 0 ..^ L − F → S substr F L ⁡ 0 = S ⁡ 0 + F
8 2 6 7 syl2anc ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → S substr F L ⁡ 0 = S ⁡ 0 + F
9 elfzoelz ⊢ F ∈ 0 ..^ L → F ∈ ℤ
10 9 zcnd ⊢ F ∈ 0 ..^ L → F ∈ ℂ
11 10 addlidd ⊢ F ∈ 0 ..^ L → 0 + F = F
12 11 fveq2d ⊢ F ∈ 0 ..^ L → S ⁡ 0 + F = S ⁡ F
13 12 3ad2ant2 ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → S ⁡ 0 + F = S ⁡ F
14 8 13 eqtrd ⊢ S ∈ Word A ∧ F ∈ 0 ..^ L ∧ L ∈ 0 … S → S substr F L ⁡ 0 = S ⁡ F