Metamath Proof Explorer


Theorem functermclem

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

Ref Expression
Hypotheses functermclem.1 ⊢ φ ∧ K R L → K = F
functermclem.2 ⊢ φ → F R L ↔ L = G
Assertion functermclem ⊢ φ → K R L ↔ K = F ∧ L = G

Proof

Step Hyp Ref Expression
1 functermclem.1 ⊢ φ ∧ K R L → K = F
2 functermclem.2 ⊢ φ → F R L ↔ L = G
3 simpr ⊢ φ ∧ K R L → K R L
4 1 3 eqbrtrrd ⊢ φ ∧ K R L → F R L
5 2 biimpa ⊢ φ ∧ F R L → L = G
6 4 5 syldan ⊢ φ ∧ K R L → L = G
7 1 6 jca ⊢ φ ∧ K R L → K = F ∧ L = G
8 simprl ⊢ φ ∧ K = F ∧ L = G → K = F
9 2 biimpar ⊢ φ ∧ L = G → F R L
10 9 adantrl ⊢ φ ∧ K = F ∧ L = G → F R L
11 8 10 eqbrtrd ⊢ φ ∧ K = F ∧ L = G → K R L
12 7 11 impbida ⊢ φ → K R L ↔ K = F ∧ L = G