let R be RelStr ; for S being Subset of R
for B being Subset of (subrelstr S)
for x being Element of (subrelstr S)
for y being Element of R st x = y & x is_minimal_in B holds
y is_minimal_in B
let S be Subset of R; for B being Subset of (subrelstr S)
for x being Element of (subrelstr S)
for y being Element of R st x = y & x is_minimal_in B holds
y is_minimal_in B
let B be Subset of (subrelstr S); for x being Element of (subrelstr S)
for y being Element of R st x = y & x is_minimal_in B holds
y is_minimal_in B
let x be Element of (subrelstr S); for y being Element of R st x = y & x is_minimal_in B holds
y is_minimal_in B
let y be Element of R; ( x = y & x is_minimal_in B implies y is_minimal_in B )
assume that
A:
x = y
and
B:
x is_minimal_in B
; y is_minimal_in B
C:
x in B
by B, WAYBEL_4:57;
assume
not y is_minimal_in B
; contradiction
then consider z being Element of R such that
A1:
z in B
and
B1:
z < y
by A, C, WAYBEL_4:57;
C1:
z <= y
by B1, ORDERS_2:def 10;
reconsider z9 = z as Element of (subrelstr S) by A1;
z9 <= x
by A1, C1, A, YELLOW_0:61;
then
z9 < x
by B1, A, ORDERS_2:def 10;
hence
contradiction
by A1, B, WAYBEL_4:57; verum