let T be non empty RelStr ; :: thesis: for A, B being Subset of T
for n being Element of NAT holds Fdfl (A /\ B),n = (Fdfl A,n) /\ (Fdfl B,n)
let A, B be Subset of T; :: thesis: for n being Element of NAT holds Fdfl (A /\ B),n = (Fdfl A,n) /\ (Fdfl B,n)
defpred S₁[ Element of NAT ] means (Fdfl (A /\ B)) . $1 = ((Fdfl A) . $1) /\ ((Fdfl B) . $1);
A1: for n being Element of NAT holds S₁[n]

proof

A2: S₁[ 0 ]

proof

(Fdfl (A /\ B)) . 0 = A /\ B by Def8
.= ((Fdfl A) . 0 ) /\ B by Def8
.= ((Fdfl A) . 0 ) /\ ((Fdfl B) . 0 ) by Def8 ;
hence S₁[ 0 ] ; :: thesis: verum

end;

A3: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A4: S₁[k] ; :: thesis: S₁[k + 1]
(Fdfl (A /\ B)) . (k + 1) = (Fdfl (A /\ B),k) ^d by Def8
.= ((Fdfl A,k) ^d ) /\ ((Fdfl B,k) ^d ) by A4, Th12
.= (Fdfl A,(k + 1)) /\ ((Fdfl B,k) ^d ) by Def8
.= ((Fdfl A) . (k + 1)) /\ ((Fdfl B) . (k + 1)) by Def8 ;
hence S₁[k + 1] ; :: thesis: verum

end;

for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A2, A3);
hence for n being Element of NAT holds S₁[n] ; :: thesis: verum

end;

let n be Element of NAT ; :: thesis: Fdfl (A /\ B),n = (Fdfl A,n) /\ (Fdfl B,n)
thus Fdfl (A /\ B),n = (Fdfl A,n) /\ (Fdfl B,n) by A1; :: thesis: verum