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Theorem prarloclemn 6655
 Description: Subtracting two from a positive integer. Lemma for prarloc 6659. (Contributed by Jim Kingdon, 5-Nov-2019.)
Assertion
Ref Expression
prarloclemn ((𝑁N ∧ 1𝑜 <N 𝑁) → ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁)
Distinct variable group:   𝑥,𝑁

Proof of Theorem prarloclemn
StepHypRef Expression
1 simpl 106 . . 3 ((𝑁N ∧ 1𝑜 <N 𝑁) → 𝑁N)
2 1pi 6471 . . . . 5 1𝑜N
3 ltpiord 6475 . . . . 5 ((1𝑜N𝑁N) → (1𝑜 <N 𝑁 ↔ 1𝑜𝑁))
42, 3mpan 408 . . . 4 (𝑁N → (1𝑜 <N 𝑁 ↔ 1𝑜𝑁))
54biimpa 284 . . 3 ((𝑁N ∧ 1𝑜 <N 𝑁) → 1𝑜𝑁)
6 piord 6467 . . . 4 (𝑁N → Ord 𝑁)
7 ordsucss 4258 . . . 4 (Ord 𝑁 → (1𝑜𝑁 → suc 1𝑜𝑁))
86, 7syl 14 . . 3 (𝑁N → (1𝑜𝑁 → suc 1𝑜𝑁))
91, 5, 8sylc 60 . 2 ((𝑁N ∧ 1𝑜 <N 𝑁) → suc 1𝑜𝑁)
10 df-2o 6033 . . . 4 2𝑜 = suc 1𝑜
1110sseq1i 2997 . . 3 (2𝑜𝑁 ↔ suc 1𝑜𝑁)
12 pinn 6465 . . . . 5 (𝑁N𝑁 ∈ ω)
13 2onn 6125 . . . . . 6 2𝑜 ∈ ω
14 nnawordex 6132 . . . . . 6 ((2𝑜 ∈ ω ∧ 𝑁 ∈ ω) → (2𝑜𝑁 ↔ ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁))
1513, 14mpan 408 . . . . 5 (𝑁 ∈ ω → (2𝑜𝑁 ↔ ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁))
1612, 15syl 14 . . . 4 (𝑁N → (2𝑜𝑁 ↔ ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁))
1716adantr 265 . . 3 ((𝑁N ∧ 1𝑜 <N 𝑁) → (2𝑜𝑁 ↔ ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁))
1811, 17syl5bbr 187 . 2 ((𝑁N ∧ 1𝑜 <N 𝑁) → (suc 1𝑜𝑁 ↔ ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁))
199, 18mpbid 139 1 ((𝑁N ∧ 1𝑜 <N 𝑁) → ∃𝑥 ∈ ω (2𝑜 +𝑜 𝑥) = 𝑁)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 101   ↔ wb 102   = wceq 1259   ∈ wcel 1409  ∃wrex 2324   ⊆ wss 2945   class class class wbr 3792  Ord word 4127  suc csuc 4130  ωcom 4341  (class class class)co 5540  1𝑜c1o 6025  2𝑜c2o 6026   +𝑜 coa 6029  Ncnpi 6428
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