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Metamath Proof Explorer

Theorem rdgsucmpt2

Description: This version of rdgsucmpt avoids the not-free hypothesis of rdgsucmptf by using two substitutions instead of one. (Contributed by Mario Carneiro, 11-Sep-2015)

Ref Expression
Hypotheses rdgsucmpt2.1 ⊢ 𝐹 = rec ( ( 𝑥 ∈ V ↦ 𝐶 ) , 𝐴 )
rdgsucmpt2.2 ⊢ ( 𝑦 = 𝑥 → 𝐸 = 𝐶 )
rdgsucmpt2.3 ⊢ ( 𝑦 = ( 𝐹 ‘ 𝐵 ) → 𝐸 = 𝐷 )
Assertion rdgsucmpt2 ( ( 𝐵 ∈ On ∧ 𝐷 ∈ 𝑉 ) → ( 𝐹 ‘ suc 𝐵 ) = 𝐷 )

Proof

Step Hyp Ref Expression
1 rdgsucmpt2.1 ⊢ 𝐹 = rec ( ( 𝑥 ∈ V ↦ 𝐶 ) , 𝐴 )
2 rdgsucmpt2.2 ⊢ ( 𝑦 = 𝑥 → 𝐸 = 𝐶 )
3 rdgsucmpt2.3 ⊢ ( 𝑦 = ( 𝐹 ‘ 𝐵 ) → 𝐸 = 𝐷 )
4 nfcv ⊢ Ⅎ 𝑦 𝐴
5 nfcv ⊢ Ⅎ 𝑦 𝐵
6 nfcv ⊢ Ⅎ 𝑦 𝐷
7 2 cbvmptv ⊢ ( 𝑦 ∈ V ↦ 𝐸 ) = ( 𝑥 ∈ V ↦ 𝐶 )
8 rdgeq1 ⊢ ( ( 𝑦 ∈ V ↦ 𝐸 ) = ( 𝑥 ∈ V ↦ 𝐶 ) → rec ( ( 𝑦 ∈ V ↦ 𝐸 ) , 𝐴 ) = rec ( ( 𝑥 ∈ V ↦ 𝐶 ) , 𝐴 ) )
9 7 8 ax-mp ⊢ rec ( ( 𝑦 ∈ V ↦ 𝐸 ) , 𝐴 ) = rec ( ( 𝑥 ∈ V ↦ 𝐶 ) , 𝐴 )
10 1 9 eqtr4i ⊢ 𝐹 = rec ( ( 𝑦 ∈ V ↦ 𝐸 ) , 𝐴 )
11 4 5 6 10 3 rdgsucmptf ⊢ ( ( 𝐵 ∈ On ∧ 𝐷 ∈ 𝑉 ) → ( 𝐹 ‘ suc 𝐵 ) = 𝐷 )