Metamath Proof Explorer


Theorem isconstr

Description: Property of being a constructible number. (Contributed by Thierry Arnoux, 19-Oct-2025)

Ref Expression
Hypothesis constr0.1 C = rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1
Assertion isconstr A Constr m ω A C m

Proof

Step Hyp Ref Expression
1 constr0.1 C = rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1
2 df-constr Constr = rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1 ω
3 1 imaeq1i C ω = rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1 ω
4 3 unieqi C ω = rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1 ω
5 2 4 eqtr4i Constr = C ω
6 5 eleq2i A Constr A C ω
7 rdgfun Fun rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1
8 1 funeqi Fun C Fun rec s V x | a s b s c s d s t r x = a + t b a x = c + r d c b a d c 0 a s b s c s e s f s t x = a + t b a x c = e f a s b s c s d s e s f s a d x a = b c x d = e f 0 1
9 7 8 mpbir Fun C
10 eluniima Fun C A C ω m ω A C m
11 9 10 ax-mp A C ω m ω A C m
12 6 11 bitri A Constr m ω A C m