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