Theorem opsqrlem3 32737

Description: Lemma for opsqri . (Contributed by NM, 22-Aug-2006.) (New usage is discouraged.)

Hypotheses

Ref	Expression
opsqrlem2.1	⊢ 𝑇 ∈ HrmOp
opsqrlem2.2	⊢ 𝑆 = (𝑥 ∈ HrmOp, 𝑦 ∈ HrmOp ↦ (𝑥 +op ((1 / 2) ·op (𝑇 −op (𝑥 ∘ 𝑥)))))
opsqrlem2.3	⊢ 𝐹 = seq1(𝑆, (ℕ × { 0hop }))

Assertion

Ref	Expression
opsqrlem3	⊢ ((𝐺 ∈ HrmOp ∧ 𝐻 ∈ HrmOp) → (𝐺𝑆𝐻) = (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))))

Distinct variable group:   𝑥,𝑦,𝑇

Allowed substitution hints:   𝑆(𝑥, 𝑦)   𝐹(𝑥, 𝑦)   𝐺(𝑥, 𝑦)   𝐻(𝑥, 𝑦)

Proof of Theorem opsqrlem3

Dummy variables 𝑧 𝑤 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		id 23	. . 3 ⊢ (𝑧 = 𝐺 → 𝑧 = 𝐺)
2	1, 1	coeq12d 5842	. . . . 5 ⊢ (𝑧 = 𝐺 → (𝑧 ∘ 𝑧) = (𝐺 ∘ 𝐺))
3	2	oveq2d 7434	. . . 4 ⊢ (𝑧 = 𝐺 → (𝑇 −op (𝑧 ∘ 𝑧)) = (𝑇 −op (𝐺 ∘ 𝐺)))
4	3	oveq2d 7434	. . 3 ⊢ (𝑧 = 𝐺 → ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧))) = ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺))))
5	1, 4	oveq12d 7436	. 2 ⊢ (𝑧 = 𝐺 → (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))) = (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))))
6		eqidd 2762	. 2 ⊢ (𝑤 = 𝐻 → (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))) = (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))))
7		opsqrlem2.2	. . 3 ⊢ 𝑆 = (𝑥 ∈ HrmOp, 𝑦 ∈ HrmOp ↦ (𝑥 +op ((1 / 2) ·op (𝑇 −op (𝑥 ∘ 𝑥)))))
8		id 23	. . . . 5 ⊢ (𝑥 = 𝑧 → 𝑥 = 𝑧)
9	8, 8	coeq12d 5842	. . . . . . 7 ⊢ (𝑥 = 𝑧 → (𝑥 ∘ 𝑥) = (𝑧 ∘ 𝑧))
10	9	oveq2d 7434	. . . . . 6 ⊢ (𝑥 = 𝑧 → (𝑇 −op (𝑥 ∘ 𝑥)) = (𝑇 −op (𝑧 ∘ 𝑧)))
11	10	oveq2d 7434	. . . . 5 ⊢ (𝑥 = 𝑧 → ((1 / 2) ·op (𝑇 −op (𝑥 ∘ 𝑥))) = ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧))))
12	8, 11	oveq12d 7436	. . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 +op ((1 / 2) ·op (𝑇 −op (𝑥 ∘ 𝑥)))) = (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))))
13		eqidd 2762	. . . 4 ⊢ (𝑦 = 𝑤 → (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))) = (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))))
14	12, 13	cbvmpov 7513	. . 3 ⊢ (𝑥 ∈ HrmOp, 𝑦 ∈ HrmOp ↦ (𝑥 +op ((1 / 2) ·op (𝑇 −op (𝑥 ∘ 𝑥))))) = (𝑧 ∈ HrmOp, 𝑤 ∈ HrmOp ↦ (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))))
15	7, 14	eqtri 2784	. 2 ⊢ 𝑆 = (𝑧 ∈ HrmOp, 𝑤 ∈ HrmOp ↦ (𝑧 +op ((1 / 2) ·op (𝑇 −op (𝑧 ∘ 𝑧)))))
16		ovex 7451	. 2 ⊢ (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))) ∈ V
17	5, 6, 15, 16	ovmpo 7578	1 ⊢ ((𝐺 ∈ HrmOp ∧ 𝐻 ∈ HrmOp) → (𝐺𝑆𝐻) = (𝐺 +op ((1 / 2) ·op (𝑇 −op (𝐺 ∘ 𝐺)))))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {csn 4584   × cxp 5649   ∘ ccom 5655  (class class class)co 7418   ∈ cmpo 7420  1c1 11194   / cdiv 11966  ℕcn 12328  2c2 12390  seqcseq 14137   +_op chos 31533   ·_op chot 31534   −_op chod 31535   0_hop ch0o 31538  HrmOpcho 31545

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-nul 5260  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-sbc 3740  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-iota 6493  df-fun 6539  df-fv 6545  df-ov 7421  df-oprab 7422  df-mpo 7423

This theorem is used by:  opsqrlem4  32738  opsqrlem5  32739