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
A1:
x = y
and
A2:
x is_minimal_in B
; y is_minimal_in B
A3:
x in B
by A2, WAYBEL_4:56;
assume
not y is_minimal_in B
; contradiction
then consider z being Element of R such that
A4:
z in B
and
A5:
z < y
by A1, A3, WAYBEL_4:56;
A6:
z <= y
by A5, ORDERS_2:def 6;
reconsider z9 = z as Element of (subrelstr S) by A4;
z9 <= x
by A4, A6, A1, YELLOW_0:60;
then
z9 < x
by A5, A1, ORDERS_2:def 6;
hence
contradiction
by A4, A2, WAYBEL_4:56; verum