Theorem elsetpreimafveq 48187

Description: If two preimages of function values contain elements with identical function values, then both preimages are equal. (Contributed by AV, 8-Mar-2024.)

Hypothesis

Ref	Expression
setpreimafvex.p	⊢ 𝑃 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = (◡𝐹 “ {(𝐹‘𝑥)})}

Assertion

Ref	Expression
elsetpreimafveq	⊢ ((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) → ((𝐹‘𝑋) = (𝐹‘𝑌) → 𝑆 = 𝑅))

Distinct variable groups:   𝑥,𝐴,𝑧   𝑥,𝐹,𝑧   𝑥,𝑅,𝑧   𝑥,𝑆,𝑧   𝑥,𝑋   𝑥,𝑌

Allowed substitution hints:   𝑃(𝑥, 𝑧)   𝑋(𝑧)   𝑌(𝑧)

Proof of Theorem elsetpreimafveq

Step	Hyp	Ref	Expression
1		eqeq2 2778	. . . . 5 ⊢ ((𝐹‘𝑋) = (𝐹‘𝑌) → ((𝐹‘𝑥) = (𝐹‘𝑋) ↔ (𝐹‘𝑥) = (𝐹‘𝑌)))
2	1	rabbidv 3426	. . . 4 ⊢ ((𝐹‘𝑋) = (𝐹‘𝑌) → {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑋)} = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑌)})
3	2	adantl 487	. . 3 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑋)} = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑌)})
4		id 23	. . . . . 6 ⊢ (𝐹 Fn 𝐴 → 𝐹 Fn 𝐴)
5		simpl 488	. . . . . 6 ⊢ ((𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) → 𝑆 ∈ 𝑃)
6		simpl 488	. . . . . 6 ⊢ ((𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅) → 𝑋 ∈ 𝑆)
7	4, 5, 6	3anim123i 1169	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) → (𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆))
8	7	adantr 486	. . . 4 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → (𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆))
9		setpreimafvex.p	. . . . 5 ⊢ 𝑃 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = (◡𝐹 “ {(𝐹‘𝑥)})}
10	9	elsetpreimafvrab 48184	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆) → 𝑆 = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑋)})
11	8, 10	syl 18	. . 3 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → 𝑆 = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑋)})
12		simpr 490	. . . . . 6 ⊢ ((𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) → 𝑅 ∈ 𝑃)
13		simpr 490	. . . . . 6 ⊢ ((𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅) → 𝑌 ∈ 𝑅)
14	4, 12, 13	3anim123i 1169	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) → (𝐹 Fn 𝐴 ∧ 𝑅 ∈ 𝑃 ∧ 𝑌 ∈ 𝑅))
15	14	adantr 486	. . . 4 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → (𝐹 Fn 𝐴 ∧ 𝑅 ∈ 𝑃 ∧ 𝑌 ∈ 𝑅))
16	9	elsetpreimafvrab 48184	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ 𝑅 ∈ 𝑃 ∧ 𝑌 ∈ 𝑅) → 𝑅 = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑌)})
17	15, 16	syl 18	. . 3 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → 𝑅 = {𝑥 ∈ 𝐴 ∣ (𝐹‘𝑥) = (𝐹‘𝑌)})
18	3, 11, 17	3eqtr4d 2811	. 2 ⊢ (((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) ∧ (𝐹‘𝑋) = (𝐹‘𝑌)) → 𝑆 = 𝑅)
19	18	ex 418	1 ⊢ ((𝐹 Fn 𝐴 ∧ (𝑆 ∈ 𝑃 ∧ 𝑅 ∈ 𝑃) ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑅)) → ((𝐹‘𝑋) = (𝐹‘𝑌) → 𝑆 = 𝑅))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2146  {cab 2744  ∃wrex 3092  {crab 3419  {csn 4594  ^◡ccnv 5665   “ cima 5669   Fn wfn 6538  ‘cfv 6543

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 2148  ax-9 2156  ax-10 2179  ax-12 2216  ax-ext 2738  ax-sep 5262  ax-nul 5274  ax-pr 5409

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 2570  df-eu 2600  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-opab 5179  df-id 5561  df-xp 5672  df-rel 5673  df-cnv 5674  df-co 5675  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-iota 6499  df-fun 6545  df-fn 6546  df-fv 6551

This theorem is used by:  imasetpreimafvbijlemf1  48194