Theorem equcomd 2019

Description: Deduction form of equcom 2018, symmetry of equality. For the versions for classes, see eqcom 2737 and eqcomd 2736. (Contributed by BJ, 6-Oct-2019.)

Hypothesis

Ref	Expression
equcomd.1	⊢ (𝜑 → 𝑥 = 𝑦)

Assertion

Ref	Expression
equcomd	⊢ (𝜑 → 𝑦 = 𝑥)

Proof of Theorem equcomd

Step	Hyp	Ref	Expression
1		equcomd.1	. 2 ⊢ (𝜑 → 𝑥 = 𝑦)
2		equcom 2018	. 2 ⊢ (𝑥 = 𝑦 ↔ 𝑦 = 𝑥)
3	1, 2	sylib 218	1 ⊢ (𝜑 → 𝑦 = 𝑥)

Syntax hints:   → wi 4

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780

This theorem is referenced by:  sndisj  5102  fsumcom2  15747  fprodcom2  15957  catideu  17643  pospo  18311  dprdfcntz  19954  ordtt1  23273  eengtrkg  28920  cusgrfilem2  29391  frgr2wwlk1  30265  ssmxidl  33452  gonar  35389  bj-nfcsym  36894  exidu1  37857  rngoideu  37904  2reu8i  47118  ichnreuop  47477  sprsymrelf1lem  47496  oppcthinendcALT  49434