Metamath Proof Explorer


Theorem sbc2rexgOLD

Description: Exchange a substitution with two existentials. (Contributed by Stefan O'Rear, 11-Oct-2014) Obsolete as of 24-Aug-2018. Use csbov123 instead. (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Assertion sbc2rexgOLD ( 𝐴𝑉 → ( [ 𝐴 / 𝑎 ]𝑏𝐵𝑐𝐶 𝜑 ↔ ∃ 𝑏𝐵𝑐𝐶 [ 𝐴 / 𝑎 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 elex ( 𝐴𝑉𝐴 ∈ V )
2 sbcrexgOLD ( 𝐴 ∈ V → ( [ 𝐴 / 𝑎 ]𝑏𝐵𝑐𝐶 𝜑 ↔ ∃ 𝑏𝐵 [ 𝐴 / 𝑎 ]𝑐𝐶 𝜑 ) )
3 sbcrexgOLD ( 𝐴 ∈ V → ( [ 𝐴 / 𝑎 ]𝑐𝐶 𝜑 ↔ ∃ 𝑐𝐶 [ 𝐴 / 𝑎 ] 𝜑 ) )
4 3 rexbidv ( 𝐴 ∈ V → ( ∃ 𝑏𝐵 [ 𝐴 / 𝑎 ]𝑐𝐶 𝜑 ↔ ∃ 𝑏𝐵𝑐𝐶 [ 𝐴 / 𝑎 ] 𝜑 ) )
5 2 4 bitrd ( 𝐴 ∈ V → ( [ 𝐴 / 𝑎 ]𝑏𝐵𝑐𝐶 𝜑 ↔ ∃ 𝑏𝐵𝑐𝐶 [ 𝐴 / 𝑎 ] 𝜑 ) )
6 1 5 syl ( 𝐴𝑉 → ( [ 𝐴 / 𝑎 ]𝑏𝐵𝑐𝐶 𝜑 ↔ ∃ 𝑏𝐵𝑐𝐶 [ 𝐴 / 𝑎 ] 𝜑 ) )