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Theorem mpt22eqb 5552
 Description: Bidirectional equality theorem for a mapping abstraction. Equivalent to eqfnov2 5550. (Contributed by Mario Carneiro, 4-Jan-2017.)
Assertion
Ref Expression
mpt22eqb (x A y B 𝐶 𝑉 → ((x A, y B𝐶) = (x A, y B𝐷) ↔ x A y B 𝐶 = 𝐷))
Distinct variable groups:   x,y,A   y,B
Allowed substitution hints:   B(x)   𝐶(x,y)   𝐷(x,y)   𝑉(x,y)

Proof of Theorem mpt22eqb
Dummy variable z is distinct from all other variables.
StepHypRef Expression
1 pm13.183 2675 . . . . . 6 (𝐶 𝑉 → (𝐶 = 𝐷z(z = 𝐶z = 𝐷)))
21ralimi 2378 . . . . 5 (y B 𝐶 𝑉y B (𝐶 = 𝐷z(z = 𝐶z = 𝐷)))
3 ralbi 2439 . . . . 5 (y B (𝐶 = 𝐷z(z = 𝐶z = 𝐷)) → (y B 𝐶 = 𝐷y B z(z = 𝐶z = 𝐷)))
42, 3syl 14 . . . 4 (y B 𝐶 𝑉 → (y B 𝐶 = 𝐷y B z(z = 𝐶z = 𝐷)))
54ralimi 2378 . . 3 (x A y B 𝐶 𝑉x A (y B 𝐶 = 𝐷y B z(z = 𝐶z = 𝐷)))
6 ralbi 2439 . . 3 (x A (y B 𝐶 = 𝐷y B z(z = 𝐶z = 𝐷)) → (x A y B 𝐶 = 𝐷x A y B z(z = 𝐶z = 𝐷)))
75, 6syl 14 . 2 (x A y B 𝐶 𝑉 → (x A y B 𝐶 = 𝐷x A y B z(z = 𝐶z = 𝐷)))
8 df-mpt2 5460 . . . 4 (x A, y B𝐶) = {⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐶)}
9 df-mpt2 5460 . . . 4 (x A, y B𝐷) = {⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐷)}
108, 9eqeq12i 2050 . . 3 ((x A, y B𝐶) = (x A, y B𝐷) ↔ {⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐶)} = {⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐷)})
11 eqoprab2b 5505 . . 3 ({⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐶)} = {⟨⟨x, y⟩, z⟩ ∣ ((x A y B) z = 𝐷)} ↔ xyz(((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)))
12 pm5.32 426 . . . . . . 7 (((x A y B) → (z = 𝐶z = 𝐷)) ↔ (((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)))
1312albii 1356 . . . . . 6 (z((x A y B) → (z = 𝐶z = 𝐷)) ↔ z(((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)))
14 19.21v 1750 . . . . . 6 (z((x A y B) → (z = 𝐶z = 𝐷)) ↔ ((x A y B) → z(z = 𝐶z = 𝐷)))
1513, 14bitr3i 175 . . . . 5 (z(((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)) ↔ ((x A y B) → z(z = 𝐶z = 𝐷)))
16152albii 1357 . . . 4 (xyz(((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)) ↔ xy((x A y B) → z(z = 𝐶z = 𝐷)))
17 r2al 2337 . . . 4 (x A y B z(z = 𝐶z = 𝐷) ↔ xy((x A y B) → z(z = 𝐶z = 𝐷)))
1816, 17bitr4i 176 . . 3 (xyz(((x A y B) z = 𝐶) ↔ ((x A y B) z = 𝐷)) ↔ x A y B z(z = 𝐶z = 𝐷))
1910, 11, 183bitri 195 . 2 ((x A, y B𝐶) = (x A, y B𝐷) ↔ x A y B z(z = 𝐶z = 𝐷))
207, 19syl6rbbr 188 1 (x A y B 𝐶 𝑉 → ((x A, y B𝐶) = (x A, y B𝐷) ↔ x A y B 𝐶 = 𝐷))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240   = wceq 1242   ∈ wcel 1390  ∀wral 2300  {coprab 5456   ↦ cmpt2 5457 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-14 1402  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-sep 3866  ax-pow 3918  ax-pr 3935  ax-setind 4220 This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-fal 1248  df-nf 1347  df-sb 1643  df-eu 1900  df-mo 1901  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ne 2203  df-ral 2305  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-pw 3353  df-sn 3373  df-pr 3374  df-op 3376  df-oprab 5459  df-mpt2 5460 This theorem is referenced by: (None)
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