Proof of Theorem tfrcllembacc
Step | Hyp | Ref
| Expression |
1 | | tfrcllembacc.3 |
. 2
⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))} |
2 | | simpr3 995 |
. . . . . . 7
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉})) |
3 | | tfrcl.f |
. . . . . . . 8
⊢ 𝐹 = recs(𝐺) |
4 | | tfrcl.g |
. . . . . . . . 9
⊢ (𝜑 → Fun 𝐺) |
5 | 4 | ad2antrr 480 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → Fun 𝐺) |
6 | | tfrcl.x |
. . . . . . . . 9
⊢ (𝜑 → Ord 𝑋) |
7 | 6 | ad2antrr 480 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → Ord 𝑋) |
8 | | simp1ll 1050 |
. . . . . . . . 9
⊢ ((((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → 𝜑) |
9 | | tfrcl.ex |
. . . . . . . . 9
⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆) |
10 | 8, 9 | syld3an1 1274 |
. . . . . . . 8
⊢ ((((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆) |
11 | | tfrcllemsucfn.1 |
. . . . . . . 8
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))} |
12 | | tfrcllembacc.4 |
. . . . . . . . 9
⊢ (𝜑 → 𝐷 ∈ 𝑋) |
13 | 12 | ad2antrr 480 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → 𝐷 ∈ 𝑋) |
14 | | simplr 520 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → 𝑧 ∈ 𝐷) |
15 | | tfrcllembacc.u |
. . . . . . . . . 10
⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋) |
16 | 15 | adantlr 469 |
. . . . . . . . 9
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋) |
17 | 16 | adantlr 469 |
. . . . . . . 8
⊢ ((((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋) |
18 | | simpr1 993 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → 𝑔:𝑧⟶𝑆) |
19 | | simpr2 994 |
. . . . . . . 8
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → 𝑔 ∈ 𝐴) |
20 | 3, 5, 7, 10, 11, 13, 14, 17, 18, 19 | tfrcllemsucaccv 6322 |
. . . . . . 7
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}) ∈ 𝐴) |
21 | 2, 20 | eqeltrd 2243 |
. . . . . 6
⊢ (((𝜑 ∧ 𝑧 ∈ 𝐷) ∧ (𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))) → ℎ ∈ 𝐴) |
22 | 21 | ex 114 |
. . . . 5
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐷) → ((𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉})) → ℎ ∈ 𝐴)) |
23 | 22 | exlimdv 1807 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐷) → (∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉})) → ℎ ∈ 𝐴)) |
24 | 23 | rexlimdva 2583 |
. . 3
⊢ (𝜑 → (∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉})) → ℎ ∈ 𝐴)) |
25 | 24 | abssdv 3216 |
. 2
⊢ (𝜑 → {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {〈𝑧, (𝐺‘𝑔)〉}))} ⊆ 𝐴) |
26 | 1, 25 | eqsstrid 3188 |
1
⊢ (𝜑 → 𝐵 ⊆ 𝐴) |