Theorem idfudiag1lem 50600

Description: Lemma for idfudiag1bas 50601 and idfudiag1 50602. (Contributed by Zhi Wang, 19-Oct-2025.)

Hypotheses

Ref	Expression
idfudiag1lem.1	⊢ (𝜑 → ( I ↾ 𝐴) = (𝐴 × {𝐵}))
idfudiag1lem.2	⊢ (𝜑 → 𝐴 ≠ ∅)

Assertion

Ref	Expression
idfudiag1lem	⊢ (𝜑 → 𝐴 = {𝐵})

Proof of Theorem idfudiag1lem

Step	Hyp	Ref	Expression
1		rnresi 6073	. . 3 ⊢ ran ( I ↾ 𝐴) = 𝐴
2		idfudiag1lem.1	. . . 4 ⊢ (𝜑 → ( I ↾ 𝐴) = (𝐴 × {𝐵}))
3	2	rneqd 5920	. . 3 ⊢ (𝜑 → ran ( I ↾ 𝐴) = ran (𝐴 × {𝐵}))
4	1, 3	eqtr3id 2810	. 2 ⊢ (𝜑 → 𝐴 = ran (𝐴 × {𝐵}))
5		idfudiag1lem.2	. . 3 ⊢ (𝜑 → 𝐴 ≠ ∅)
6		rnxp 6162	. . 3 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × {𝐵}) = {𝐵})
7	5, 6	syl 18	. 2 ⊢ (𝜑 → ran (𝐴 × {𝐵}) = {𝐵})
8	4, 7	eqtrd 2796	1 ⊢ (𝜑 → 𝐴 = {𝐵})

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ≠ wne 2956  ∅c0 4279  {csn 4584   I cid 5545   × cxp 5649  ran crn 5652   ↾ cres 5653

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-ext 2733  ax-sep 5249  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-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  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-br 5104  df-opab 5168  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664

This theorem is used by:  idfudiag1bas  50601  idfudiag1  50602