let n, k, m be Element of NAT ; :: thesis: for f being Function of n,k
for g being Function of (n + 1),(k + m) st f is onto & f is "increasing & f = g | n holds
for i, j being Element of NAT st i in rng g & j in rng g & i < j holds
min* (g " {i}) < min* (g " {j})
let f be Function of n,k; :: thesis: for g being Function of (n + 1),(k + m) st f is onto & f is "increasing & f = g | n holds
for i, j being Element of NAT st i in rng g & j in rng g & i < j holds
min* (g " {i}) < min* (g " {j})
let g be Function of (n + 1),(k + m); :: thesis: ( f is onto & f is "increasing & f = g | n implies for i, j being Element of NAT st i in rng g & j in rng g & i < j holds
min* (g " {i}) < min* (g " {j}) )
assume that
A1:
( f is onto & f is "increasing )
and
A2:
f = g | n
; :: thesis: for i, j being Element of NAT st i in rng g & j in rng g & i < j holds
min* (g " {i}) < min* (g " {j})
A3:
for i being Element of NAT st i < n holds
f . i = g . i
let i, j be Element of NAT ; :: thesis: ( i in rng g & j in rng g & i < j implies min* (g " {i}) < min* (g " {j}) )
assume A5:
( i in rng g & j in rng g & i < j )
; :: thesis: min* (g " {i}) < min* (g " {j})
A6:
for l being Element of NAT st l in rng g & not l in rng f holds
l = g . n
proof
let l be
Element of
NAT ;
:: thesis: ( l in rng g & not l in rng f implies l = g . n )
assume A7:
(
l in rng g & not
l in rng f )
;
:: thesis: l = g . n
assume A8:
l <> g . n
;
:: thesis: contradiction
consider x being
set such that A9:
(
x in dom g &
g . x = l )
by A7, FUNCT_1:def 5;
(
dom g = n + 1 &
n + 1 is
Subset of
NAT )
by A7, Th8, FUNCT_2:def 1;
then reconsider x =
x as
Element of
NAT by A9;
x < n + 1
by A9, NAT_1:45;
then
x <= n
by NAT_1:13;
then A10:
x < n
by A8, A9, XXREAL_0:1;
then
k <> 0
by A1, Def1;
then
(
dom f = n &
x in n &
f . x = g . x )
by A3, A10, FUNCT_2:def 1, NAT_1:45;
hence
contradiction
by A7, A9, FUNCT_1:def 5;
:: thesis: verum
end;
A11:
for l being Element of NAT st l in rng g & not l in rng f holds
min* (g " {l}) = n
proof
let l be
Element of
NAT ;
:: thesis: ( l in rng g & not l in rng f implies min* (g " {l}) = n )
assume A12:
(
l in rng g & not
l in rng f )
;
:: thesis: min* (g " {l}) = n
assume A13:
min* (g " {l}) <> n
;
:: thesis: contradiction
(
n < n + 1 &
dom g = n + 1 )
by A12, FUNCT_2:def 1, NAT_1:13;
then
(
n + 1
> 0 &
n in dom g &
l in {l} &
g . n = l )
by A6, A12, NAT_1:45, TARSKI:def 1;
then
(
min* (g " {l}) <= (n + 1) - 1 &
n in g " {l} )
by Th16, FUNCT_1:def 13;
then A14:
(
min* (g " {l}) < n &
min* (g " {l}) in g " {l} )
by A13, NAT_1:def 1, XXREAL_0:1;
then A15:
(
min* (g " {l}) in n &
g . (min* (g " {l})) in {l} &
0 < n &
f . (min* (g " {l})) = g . (min* (g " {l})) )
by A3, FUNCT_1:def 13, NAT_1:45;
k <> 0
by A1, A14, Def1;
then
dom f = n
by FUNCT_2:def 1;
then
f . (min* (g " {l})) in rng f
by A15, FUNCT_1:def 5;
hence
contradiction
by A12, A15, TARSKI:def 1;
:: thesis: verum
end;
A16:
for l being Element of NAT st l in rng g & not l in rng f holds
l >= k
A18:
for k1 being Element of NAT st k1 in rng f holds
min* (g " {k1}) = min* (f " {k1})
proof
let k1 be
Element of
NAT ;
:: thesis: ( k1 in rng f implies min* (g " {k1}) = min* (f " {k1}) )
assume A19:
k1 in rng f
;
:: thesis: min* (g " {k1}) = min* (f " {k1})
consider x being
set such that A20:
(
x in dom f &
f . x = k1 )
by A19, FUNCT_1:def 5;
k <= k + m
by NAT_1:11;
then
k c= k + m
by NAT_1:40;
then
(
k1 in k &
k1 in k + m &
n <= n + 1 )
by A19, NAT_1:11, TARSKI:def 3;
then A21:
(
dom f = n &
dom g = n + 1 &
n c= n + 1 )
by FUNCT_2:def 1, NAT_1:40;
(
x in n &
n is
Subset of
NAT )
by A20, Th8;
then reconsider x =
x as
Element of
NAT ;
x < n
by A20, NAT_1:45;
then
(
f . x = g . x &
f . x = k1 &
k1 in {k1} &
x in dom g )
by A3, A20, A21, TARSKI:def 1;
then
(
x in f " {k1} &
x in g " {k1} &
k1 in rng g )
by A20, FUNCT_1:def 5, FUNCT_1:def 13;
then A22:
(
min* (f " {k1}) <= x &
x < n &
min* (f " {k1}) in f " {k1} )
by A20, NAT_1:45, NAT_1:def 1;
then
(
min* (f " {k1}) < n &
f . (min* (f " {k1})) in {k1} )
by FUNCT_1:def 13, XXREAL_0:2;
then
(
min* (f " {k1}) in n &
g . (min* (f " {k1})) in {k1} )
by A3, NAT_1:45;
then A23:
min* (f " {k1}) in g " {k1}
by A21, FUNCT_1:def 13;
hence
min* (g " {k1}) = min* (f " {k1})
by A23, NAT_1:def 1;
:: thesis: verum
end;
hence
min* (g " {i}) < min* (g " {j})
; :: thesis: verum