Theorem equid1 2158

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 (although the proof, as you can see below, is not as obvious as you might think). This proof uses only axioms without distinct variable conditions and thus requires no dummy variables. A simpler proof, similar to Tarki's, is possible if we make use of ax-17 1616; see the proof of equid 1676. See equid1ALT 2176 for an alternate proof. (Contributed by NM, 5-Aug-1993.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
equid1	⊢ x = x

Proof of Theorem equid1

Step	Hyp	Ref	Expression
1		ax-5o 2136	. . . 4 ⊢ (∀x(∀x ¬ ∀x x = x → (x = x → ∀x x = x)) → (∀x ¬ ∀x x = x → ∀x(x = x → ∀x x = x)))
2		ax-4 2135	. . . . 5 ⊢ (∀x ¬ ∀x x = x → ¬ ∀x x = x)
3		ax-12o 2142	. . . . 5 ⊢ (¬ ∀x x = x → (¬ ∀x x = x → (x = x → ∀x x = x)))
4	2, 2, 3	sylc 56	. . . 4 ⊢ (∀x ¬ ∀x x = x → (x = x → ∀x x = x))
5	1, 4	mpg 1548	. . 3 ⊢ (∀x ¬ ∀x x = x → ∀x(x = x → ∀x x = x))
6		ax-9o 2138	. . 3 ⊢ (∀x(x = x → ∀x x = x) → x = x)
7	5, 6	syl 15	. 2 ⊢ (∀x ¬ ∀x x = x → x = x)
8		ax-6o 2137	. 2 ⊢ (¬ ∀x ¬ ∀x x = x → x = x)
9	7, 8	pm2.61i 156	1 ⊢ x = x

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-4 2135  ax-5o 2136  ax-6o 2137  ax-9o 2138  ax-12o 2142

This theorem is referenced by: (None)