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Theorem equcomi 1990
Description: Commutative law for equality. Equality is a symmetric relation. Lemma 3 of [KalishMontague] p. 85. See also Lemma 7 of [Tarski] p. 69. (Contributed by NM, 10-Jan-1993.) (Revised by NM, 9-Apr-2017.)
Assertion
Ref Expression
equcomi (𝑥 = 𝑦𝑦 = 𝑥)

Proof of Theorem equcomi
StepHypRef Expression
1 equid 1985 . 2 𝑥 = 𝑥
2 ax7 1989 . 2 (𝑥 = 𝑦 → (𝑥 = 𝑥𝑦 = 𝑥))
31, 2mpi 20 1 (𝑥 = 𝑦𝑦 = 𝑥)
Colors of variables: wff setvar class
Syntax hints:  wi 4
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981
This theorem depends on definitions:  df-bi 197  df-an 385  df-ex 1745
This theorem is referenced by:  equcom  1991  equcoms  1993  ax13dgen2  2055  cbv2h  2305  axc11nOLD  2343  axc16i  2353  equsb2  2397  axsep  4813  rext  4946  soxp  7335  axextnd  9451  prodmo  14710  mpt2matmul  20300  finminlem  32437  bj-ssbid2ALT  32771  axc11n11  32797  axc11n11r  32798  bj-cbv2hv  32856  bj-axsep  32918  ax6er  32945  poimirlem25  33564  axc11nfromc11  34530  aev-o  34535
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