Metamath Proof Explorer


Theorem fvmptd

Description: Deduction version of fvmpt . (Contributed by Scott Fenton, 18-Feb-2013) (Revised by Mario Carneiro, 31-Aug-2015) (Proof shortened by AV, 29-Mar-2024)

Ref Expression
Hypotheses fvmptd.1 φ F = x D B
fvmptd.2 φ x = A B = C
fvmptd.3 φ A D
fvmptd.4 φ C V
Assertion fvmptd φ F A = C

Proof

Step Hyp Ref Expression
1 fvmptd.1 φ F = x D B
2 fvmptd.2 φ x = A B = C
3 fvmptd.3 φ A D
4 fvmptd.4 φ C V
5 nfv x φ
6 nfcv _ x A
7 nfcv _ x C
8 1 2 3 4 5 6 7 fvmptdf φ F A = C