Theorem functermclem 50342

Description: Lemma for functermc 50343. (Contributed by Zhi Wang, 17-Oct-2025.)

Hypotheses

Ref	Expression
functermclem.1	⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → 𝐾 = 𝐹)
functermclem.2	⊢ (𝜑 → (𝐹𝑅𝐿 ↔ 𝐿 = 𝐺))

Assertion

Ref	Expression
functermclem	⊢ (𝜑 → (𝐾𝑅𝐿 ↔ (𝐾 = 𝐹 ∧ 𝐿 = 𝐺)))

Proof of Theorem functermclem

Step	Hyp	Ref	Expression
1		functermclem.1	. . 3 ⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → 𝐾 = 𝐹)
2		simpr 490	. . . . 5 ⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → 𝐾𝑅𝐿)
3	1, 2	eqbrtrrd 5137	. . . 4 ⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → 𝐹𝑅𝐿)
4		functermclem.2	. . . . 5 ⊢ (𝜑 → (𝐹𝑅𝐿 ↔ 𝐿 = 𝐺))
5	4	biimpa 482	. . . 4 ⊢ ((𝜑 ∧ 𝐹𝑅𝐿) → 𝐿 = 𝐺)
6	3, 5	syldan 603	. . 3 ⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → 𝐿 = 𝐺)
7	1, 6	jca 521	. 2 ⊢ ((𝜑 ∧ 𝐾𝑅𝐿) → (𝐾 = 𝐹 ∧ 𝐿 = 𝐺))
8		simprl 783	. . 3 ⊢ ((𝜑 ∧ (𝐾 = 𝐹 ∧ 𝐿 = 𝐺)) → 𝐾 = 𝐹)
9	4	biimpar 483	. . . 4 ⊢ ((𝜑 ∧ 𝐿 = 𝐺) → 𝐹𝑅𝐿)
10	9	adantrl 729	. . 3 ⊢ ((𝜑 ∧ (𝐾 = 𝐹 ∧ 𝐿 = 𝐺)) → 𝐹𝑅𝐿)
11	8, 10	eqbrtrd 5135	. 2 ⊢ ((𝜑 ∧ (𝐾 = 𝐹 ∧ 𝐿 = 𝐺)) → 𝐾𝑅𝐿)
12	7, 11	impbida 813	1 ⊢ (𝜑 → (𝐾𝑅𝐿 ↔ (𝐾 = 𝐹 ∧ 𝐿 = 𝐺)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   class class class wbr 5111

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112

This theorem is used by:  functermc  50343