Metamath Proof Explorer


Theorem idfudiag1lem

Description: Lemma for idfudiag1bas and idfudiag1 . (Contributed by Zhi Wang, 19-Oct-2025)

Ref Expression
Hypotheses idfudiag1lem.1 ⊢ φ → I ↾ A = A × B
idfudiag1lem.2 ⊢ φ → A ≠ ∅
Assertion idfudiag1lem ⊢ φ → A = B

Proof

Step Hyp Ref Expression
1 idfudiag1lem.1 ⊢ φ → I ↾ A = A × B
2 idfudiag1lem.2 ⊢ φ → A ≠ ∅
3 rnresi ⊢ ran ⁡ I ↾ A = A
4 1 rneqd ⊢ φ → ran ⁡ I ↾ A = ran ⁡ A × B
5 3 4 eqtr3id ⊢ φ → A = ran ⁡ A × B
6 rnxp ⊢ A ≠ ∅ → ran ⁡ A × B = B
7 2 6 syl ⊢ φ → ran ⁡ A × B = B
8 5 7 eqtrd ⊢ φ → A = B