Theorem equid 1605

Description: Identity law for equality (reflexivity). Lemma 6 of [Tarski] p. 68. This is often an axiom of equality in textbook systems, but we don't need it as an axiom since it can be proved from our other axioms.

This proof is similar to Tarski's and makes use of a dummy variable 𝑦. It also works in intuitionistic logic, unlike some other possible ways of proving this theorem. (Contributed by NM, 1-Apr-2005.)

Assertion

Ref	Expression
equid	⊢ 𝑥 = 𝑥

Proof of Theorem equid

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		a9e 1602	. 2 ⊢ ∃𝑦 𝑦 = 𝑥
2		ax-17 1435	. . 3 ⊢ (𝑥 = 𝑥 → ∀𝑦 𝑥 = 𝑥)
3		ax-8 1411	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 = 𝑥 → 𝑥 = 𝑥))
4	3	pm2.43i 47	. . 3 ⊢ (𝑦 = 𝑥 → 𝑥 = 𝑥)
5	2, 4	exlimih 1500	. 2 ⊢ (∃𝑦 𝑦 = 𝑥 → 𝑥 = 𝑥)
6	1, 5	ax-mp 7	1 ⊢ 𝑥 = 𝑥

Syntax hints:  ∃wex 1397

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-gen 1354  ax-ie2 1399  ax-8 1411  ax-17 1435  ax-i9 1439

This theorem depends on definitions:  df-bi 114

This theorem is referenced by:  nfequid  1606  stdpc6  1607  equcomi  1608  equveli  1658  sbid  1673  ax16i  1754  exists1  2012  vjust  2575  vex  2577  reu6  2753  nfccdeq  2785  sbc8g  2794  dfnul3  3255  rab0  3274  int0  3657  ruv  4302  relop  4514  f1eqcocnv  5459  mpt2xopoveq  5886