Theorem oprabval2g 3050

Description: The value of an operation abstraction. Special case.

Hypotheses

Ref	Expression
oprabval2g.1	⊢ (x = A → R = G)
oprabval2g.2	⊢ (y = B → G = S)
oprabval2g.3	⊢ F = {⟨⟨x, y⟩, z⟩∣((x ∈ C ∧ y ∈ D) ∧ z = R)}

Assertion

Ref	Expression
oprabval2g	⊢ ((A ∈ C ∧ B ∈ D ∧ S ∈ H) → (AFB) = S)

Distinct variable group(s):   x,y,z,A   x,B,y,z   x,C,y,z   x,D,y,z   x,H,y,z   x,G   z,R   x,S,y,z

Proof of Theorem oprabval2g

Step	Hyp	Ref	Expression
1		cleqid 1102	. 2 ⊢ S = S
2		oprabval2g.1	. . . 4 ⊢ (x = A → R = G)
3	2	cleq2d 1112	. . 3 ⊢ (x = A → (z = R ↔ z = G))
4		oprabval2g.2	. . . 4 ⊢ (y = B → G = S)
5	4	cleq2d 1112	. . 3 ⊢ (y = B → (z = G ↔ z = S))
6		cleq1 1107	. . 3 ⊢ (z = S → (z = S ↔ S = S))
7		moeq 1431	. . . 4 ⊢ ∃*z z = R
8	7	a1i 7	. . 3 ⊢ ((x ∈ C ∧ y ∈ D) → ∃*z z = R)
9		oprabval2g.3	. . 3 ⊢ F = {⟨⟨x, y⟩, z⟩∣((x ∈ C ∧ y ∈ D) ∧ z = R)}
10	3, 5, 6, 8, 9	oprabvalig 3048	. 2 ⊢ ((A ∈ C ∧ B ∈ D ∧ S ∈ H) → (S = S → (AFB) = S))
11	1, 10	mpi 44	1 ⊢ ((A ∈ C ∧ B ∈ D ∧ S ∈ H) → (AFB) = S)

Syntax hints:   → wi 2   ∧ wa 196   ∧ w3a 581  ∃*wmo 1008   = wceq 1091   ∈ wcel 1092  (class class class)co 3001  {copab2 3002

This theorem is referenced by:  oprabval2 3051  oprabval4g 3053  mapvalg 3263  cdavalt 3716

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3an 583  df-ex 679  df-sb 853  df-eu 1009  df-mo 1010  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-uni 1920  df-br 2063  df-opab 2098  df-id 2125  df-xp 2424  df-rel 2425  df-cnv 2426  df-co 2427  df-dm 2428  df-rn 2429  df-res 2430  df-ima 2431  df-fun 2432  df-fv 2438  df-opr 3003  df-oprab 3004

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