Theorem frege55b 44896

Description: Lemma for frege57b 44898. Proposition 55 of [Frege1879] p. 50.

Note that eqtr2 2782 incorporates eqcom 2768 which is stronger than this proposition which is identical to equcomi 2050. Is it possible that Frege tricked himself into assuming what he was out to prove? (Contributed by RP, 24-Dec-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
frege55b	⊢ (𝑥 = 𝑦 → 𝑦 = 𝑥)

Proof of Theorem frege55b

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frege55lem2b 44895	. 2 ⊢ (𝑥 = 𝑦 → [𝑦 / 𝑧]𝑧 = 𝑥)
2		dfsb1 2511	. . 3 ⊢ ([𝑦 / 𝑧]𝑧 = 𝑥 ↔ ((𝑧 = 𝑦 → 𝑧 = 𝑥) ∧ ∃𝑧(𝑧 = 𝑦 ∧ 𝑧 = 𝑥)))
3		eqtr2 2782	. . . . 5 ⊢ ((𝑧 = 𝑦 ∧ 𝑧 = 𝑥) → 𝑦 = 𝑥)
4	3	exlimiv 1963	. . . 4 ⊢ (∃𝑧(𝑧 = 𝑦 ∧ 𝑧 = 𝑥) → 𝑦 = 𝑥)
5	4	adantl 487	. . 3 ⊢ (((𝑧 = 𝑦 → 𝑧 = 𝑥) ∧ ∃𝑧(𝑧 = 𝑦 ∧ 𝑧 = 𝑥)) → 𝑦 = 𝑥)
6	2, 5	sylbi 220	. 2 ⊢ ([𝑦 / 𝑧]𝑧 = 𝑥 → 𝑦 = 𝑥)
7	1, 6	syl 18	1 ⊢ (𝑥 = 𝑦 → 𝑦 = 𝑥)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∃wex 1812  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-13 2402  ax-ext 2733  ax-frege8 44808  ax-frege52c 44887

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-sbc 3740

This theorem is used by:  frege56b  44897