Metamath Proof Explorer


Theorem cdleme31fv1s

Description: Part of proof of Lemma E in Crawley p. 113. (Contributed by NM, 25-Feb-2013)

Ref Expression
Hypotheses cdleme31.o 𝑂 = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑥 𝑊 ) ) = 𝑥 ) → 𝑧 = ( 𝑁 ( 𝑥 𝑊 ) ) ) )
cdleme31.f 𝐹 = ( 𝑥𝐵 ↦ if ( ( 𝑃𝑄 ∧ ¬ 𝑥 𝑊 ) , 𝑂 , 𝑥 ) )
Assertion cdleme31fv1s ( ( 𝑋𝐵 ∧ ( 𝑃𝑄 ∧ ¬ 𝑋 𝑊 ) ) → ( 𝐹𝑋 ) = 𝑋 / 𝑥 𝑂 )

Proof

Step Hyp Ref Expression
1 cdleme31.o 𝑂 = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑥 𝑊 ) ) = 𝑥 ) → 𝑧 = ( 𝑁 ( 𝑥 𝑊 ) ) ) )
2 cdleme31.f 𝐹 = ( 𝑥𝐵 ↦ if ( ( 𝑃𝑄 ∧ ¬ 𝑥 𝑊 ) , 𝑂 , 𝑥 ) )
3 eqid ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑋 𝑊 ) ) = 𝑋 ) → 𝑧 = ( 𝑁 ( 𝑋 𝑊 ) ) ) ) = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑋 𝑊 ) ) = 𝑋 ) → 𝑧 = ( 𝑁 ( 𝑋 𝑊 ) ) ) )
4 1 2 3 cdleme31fv1 ( ( 𝑋𝐵 ∧ ( 𝑃𝑄 ∧ ¬ 𝑋 𝑊 ) ) → ( 𝐹𝑋 ) = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑋 𝑊 ) ) = 𝑋 ) → 𝑧 = ( 𝑁 ( 𝑋 𝑊 ) ) ) ) )
5 1 3 cdleme31so ( 𝑋𝐵 𝑋 / 𝑥 𝑂 = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑋 𝑊 ) ) = 𝑋 ) → 𝑧 = ( 𝑁 ( 𝑋 𝑊 ) ) ) ) )
6 5 adantr ( ( 𝑋𝐵 ∧ ( 𝑃𝑄 ∧ ¬ 𝑋 𝑊 ) ) → 𝑋 / 𝑥 𝑂 = ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑋 𝑊 ) ) = 𝑋 ) → 𝑧 = ( 𝑁 ( 𝑋 𝑊 ) ) ) ) )
7 4 6 eqtr4d ( ( 𝑋𝐵 ∧ ( 𝑃𝑄 ∧ ¬ 𝑋 𝑊 ) ) → ( 𝐹𝑋 ) = 𝑋 / 𝑥 𝑂 )