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Metamath Proof Explorer

Theorem cdleme31sn1c

Description: Part of proof of Lemma E in Crawley p. 113. (Contributed by NM, 1-Mar-2013)

Ref Expression
Hypotheses cdleme31sn1c.g ⊢ 𝐺 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝐸 ∨ ( ( 𝑠 ∨ 𝑡 ) ∧ 𝑊 ) ) )
cdleme31sn1c.i ⊢ 𝐼 = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝐺 ) )
cdleme31sn1c.n ⊢ 𝑁 = if ( 𝑠 ≤ ( 𝑃 ∨ 𝑄 ) , 𝐼 , 𝐷 )
cdleme31sn1c.y ⊢ 𝑌 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝐸 ∨ ( ( 𝑅 ∨ 𝑡 ) ∧ 𝑊 ) ) )
cdleme31sn1c.c ⊢ 𝐶 = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝑌 ) )
Assertion cdleme31sn1c ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ⦋ 𝑅 / 𝑠 ⦌ 𝑁 = 𝐶 )

Proof

Step Hyp Ref Expression
1 cdleme31sn1c.g ⊢ 𝐺 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝐸 ∨ ( ( 𝑠 ∨ 𝑡 ) ∧ 𝑊 ) ) )
2 cdleme31sn1c.i ⊢ 𝐼 = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝐺 ) )
3 cdleme31sn1c.n ⊢ 𝑁 = if ( 𝑠 ≤ ( 𝑃 ∨ 𝑄 ) , 𝐼 , 𝐷 )
4 cdleme31sn1c.y ⊢ 𝑌 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝐸 ∨ ( ( 𝑅 ∨ 𝑡 ) ∧ 𝑊 ) ) )
5 cdleme31sn1c.c ⊢ 𝐶 = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝑌 ) )
6 eqid ⊢ ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ) = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) )
7 2 3 6 cdleme31sn1 ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ⦋ 𝑅 / 𝑠 ⦌ 𝑁 = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ) )
8 1 4 cdleme31se ⊢ ( 𝑅 ∈ 𝐴 → ⦋ 𝑅 / 𝑠 ⦌ 𝐺 = 𝑌 )
9 8 adantr ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ⦋ 𝑅 / 𝑠 ⦌ 𝐺 = 𝑌 )
10 9 eqeq2d ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ( 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ↔ 𝑦 = 𝑌 ) )
11 10 imbi2d ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ( ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ↔ ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝑌 ) ) )
12 11 ralbidv ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ( ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ↔ ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝑌 ) ) )
13 12 riotabidv ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ) = ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = 𝑌 ) ) )
14 13 5 eqtr4di ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ( ℩ 𝑦 ∈ 𝐵 ∀ 𝑡 ∈ 𝐴 ( ( ¬ 𝑡 ≤ 𝑊 ∧ ¬ 𝑡 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑦 = ⦋ 𝑅 / 𝑠 ⦌ 𝐺 ) ) = 𝐶 )
15 7 14 eqtrd ⊢ ( ( 𝑅 ∈ 𝐴 ∧ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ⦋ 𝑅 / 𝑠 ⦌ 𝑁 = 𝐶 )