Metamath Proof Explorer


Theorem sb5f

Description: Equivalence for substitution when y is not free in ph . The implication "to the right" is sb1 and does not require the non-freeness hypothesis. Theorem sb5 replaces the non-freeness hypothesis with a disjoint variable condition and uses less axioms. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) (Revised by Mario Carneiro, 4-Oct-2016) (New usage is discouraged.)

Ref Expression
Hypothesis sb6f.1
|- F/ y ph
Assertion sb5f
|- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )

Proof

Step Hyp Ref Expression
1 sb6f.1
 |-  F/ y ph
2 1 sb6f
 |-  ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
3 1 equs45f
 |-  ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )
4 2 3 bitr4i
 |-  ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )