Metamath Proof Explorer


Theorem axlowdimlem11

Description: Lemma for axlowdim . Calculate the value of Q at its distinguished point. (Contributed by Scott Fenton, 21-Apr-2013)

Ref Expression
Hypothesis axlowdimlem10.1 ⊢ Q = I + 1 1 ∪ 1 … N ∖ I + 1 × 0
Assertion axlowdimlem11 ⊢ Q ⁡ I + 1 = 1

Proof

Step Hyp Ref Expression
1 axlowdimlem10.1 ⊢ Q = I + 1 1 ∪ 1 … N ∖ I + 1 × 0
2 1 fveq1i ⊢ Q ⁡ I + 1 = I + 1 1 ∪ 1 … N ∖ I + 1 × 0 ⁡ I + 1
3 ovex ⊢ I + 1 ∈ V
4 1ex ⊢ 1 ∈ V
5 3 4 fnsn ⊢ I + 1 1 Fn I + 1
6 c0ex ⊢ 0 ∈ V
7 6 fconst ⊢ 1 … N ∖ I + 1 × 0 : 1 … N ∖ I + 1 ⟶ 0
8 ffn ⊢ 1 … N ∖ I + 1 × 0 : 1 … N ∖ I + 1 ⟶ 0 → 1 … N ∖ I + 1 × 0 Fn 1 … N ∖ I + 1
9 7 8 ax-mp ⊢ 1 … N ∖ I + 1 × 0 Fn 1 … N ∖ I + 1
10 disjdif ⊢ I + 1 ∩ 1 … N ∖ I + 1 = ∅
11 3 snid ⊢ I + 1 ∈ I + 1
12 10 11 pm3.2i ⊢ I + 1 ∩ 1 … N ∖ I + 1 = ∅ ∧ I + 1 ∈ I + 1
13 fvun1 ⊢ I + 1 1 Fn I + 1 ∧ 1 … N ∖ I + 1 × 0 Fn 1 … N ∖ I + 1 ∧ I + 1 ∩ 1 … N ∖ I + 1 = ∅ ∧ I + 1 ∈ I + 1 → I + 1 1 ∪ 1 … N ∖ I + 1 × 0 ⁡ I + 1 = I + 1 1 ⁡ I + 1
14 5 9 12 13 mp3an ⊢ I + 1 1 ∪ 1 … N ∖ I + 1 × 0 ⁡ I + 1 = I + 1 1 ⁡ I + 1
15 3 4 fvsn ⊢ I + 1 1 ⁡ I + 1 = 1
16 2 14 15 3eqtri ⊢ Q ⁡ I + 1 = 1