Theorem functermclem 50749

Description: Lemma for functermc 50750. (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 5128	. . . 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 5126	. 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 1569   class class class wbr 5102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103

This theorem is used by:  functermc  50750