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Metamath Proof Explorer

Theorem sb6f

Description: Equivalence for substitution when y is not free in ph . The implication "to the left" is sb2 and does not require the nonfreeness hypothesis. Theorem sb6 replaces the nonfreeness hypothesis with a disjoint variable condition on x , y and requires fewer axioms. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 2-Jun-1993) (Revised by Mario Carneiro, 4-Oct-2016) (New usage is discouraged.)

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

Proof

Step Hyp Ref Expression
1 sb6f.1 
 |-  F/ y ph
2 1 nf5ri 
 |-  ( ph -> A. y ph )
3 2 sbimi 
 |-  ( [ y / x ] ph -> [ y / x ] A. y ph )
4 sb4a 
 |-  ( [ y / x ] A. y ph -> A. x ( x = y -> ph ) )
5 3 4 syl 
 |-  ( [ y / x ] ph -> A. x ( x = y -> ph ) )
6 sb2 
 |-  ( A. x ( x = y -> ph ) -> [ y / x ] ph )
7 5 6 impbii 
 |-  ( [ y / x ] ph <-> A. x ( x = y -> ph ) )