Theorem rexraleqim 3600

Description: Statement following from existence and generalization with equality. (Contributed by AV, 9-Feb-2019.)

Hypotheses

Ref	Expression
rexraleqim.1	⊢ (𝑥 = 𝑧 → (𝜓 ↔ 𝜑))
rexraleqim.2	⊢ (𝑧 = 𝑌 → (𝜑 ↔ 𝜃))

Assertion

Ref	Expression
rexraleqim	⊢ ((∃𝑧 ∈ 𝐴 𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌)) → 𝜃)

Distinct variable groups:   𝑥,𝐴,𝑧   𝑥,𝑌,𝑧   𝜑,𝑥   𝜓,𝑧   𝜃,𝑧

Allowed substitution hints:   𝜑(𝑧)   𝜓(𝑥)   𝜃(𝑥)

Proof of Theorem rexraleqim

Step	Hyp	Ref	Expression
1		rexraleqim.1	. . . . . . 7 ⊢ (𝑥 = 𝑧 → (𝜓 ↔ 𝜑))
2		eqeq1 2764	. . . . . . 7 ⊢ (𝑥 = 𝑧 → (𝑥 = 𝑌 ↔ 𝑧 = 𝑌))
3	1, 2	imbi12d 347	. . . . . 6 ⊢ (𝑥 = 𝑧 → ((𝜓 → 𝑥 = 𝑌) ↔ (𝜑 → 𝑧 = 𝑌)))
4	3	rspcva 3574	. . . . 5 ⊢ ((𝑧 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌)) → (𝜑 → 𝑧 = 𝑌))
5		rexraleqim.2	. . . . . 6 ⊢ (𝑧 = 𝑌 → (𝜑 ↔ 𝜃))
6	5	biimpd 232	. . . . 5 ⊢ (𝑧 = 𝑌 → (𝜑 → 𝜃))
7	4, 6	syli 40	. . . 4 ⊢ ((𝑧 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌)) → (𝜑 → 𝜃))
8	7	impancom 457	. . 3 ⊢ ((𝑧 ∈ 𝐴 ∧ 𝜑) → (∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌) → 𝜃))
9	8	rexlimiva 3155	. 2 ⊢ (∃𝑧 ∈ 𝐴 𝜑 → (∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌) → 𝜃))
10	9	imp 412	1 ⊢ ((∃𝑧 ∈ 𝐴 𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝑥 = 𝑌)) → 𝜃)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  ∀wral 3076  ∃wrex 3086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087

This theorem is used by:  cramerlem3  23013  elttcirr  37412