Theorem equcomd 1700

Description: Deduction form of equcom 1699, symmetry of equality. For the versions for classes, see eqcom 2172 and eqcomd 2176. (Contributed by BJ, 6-Oct-2019.)

Hypothesis

Ref	Expression
equcomd.1	|- ( ph -> x = y )

Assertion

Ref	Expression
equcomd	|- ( ph -> y = x )

Proof of Theorem equcomd

Step	Hyp	Ref	Expression
1		equcomd.1	. 2 |- ( ph -> x = y )
2		equcom 1699	. 2 |- ( x = y <-> y = x )
3	1, 2	sylib 121	1 |- ( ph -> y = x )

Syntax hints:   -> wi 4

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-gen 1442  ax-ie2 1487  ax-8 1497  ax-17 1519  ax-i9 1523

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  fisumcom2  11401  fprodcom2fi  11589  trirec0  14076