Metamath Proof Explorer


Theorem opidon2OLD

Description: Obsolete version of mgmfod as of 23-Jan-2020. An operation with a left and right identity element is onto. (Contributed by FL, 2-Nov-2009) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Hypothesis opidon2OLD.1 ⊢ X = ran ⁡ G
Assertion opidon2OLD ⊢ G ∈ Magma ∩ ExId → G : X × X ⟶ onto X

Proof

Step Hyp Ref Expression
1 opidon2OLD.1 ⊢ X = ran ⁡ G
2 eqid ⊢ dom ⁡ dom ⁡ G = dom ⁡ dom ⁡ G
3 2 opidonOLD ⊢ G ∈ Magma ∩ ExId → G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G
4 forn ⊢ G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G → ran ⁡ G = dom ⁡ dom ⁡ G
5 1 4 eqtr2id ⊢ G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G → dom ⁡ dom ⁡ G = X
6 xpeq12 ⊢ dom ⁡ dom ⁡ G = X ∧ dom ⁡ dom ⁡ G = X → dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G = X × X
7 6 anidms ⊢ dom ⁡ dom ⁡ G = X → dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G = X × X
8 foeq2 ⊢ dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G = X × X → G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G ↔ G : X × X ⟶ onto dom ⁡ dom ⁡ G
9 7 8 syl ⊢ dom ⁡ dom ⁡ G = X → G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G ↔ G : X × X ⟶ onto dom ⁡ dom ⁡ G
10 foeq3 ⊢ dom ⁡ dom ⁡ G = X → G : X × X ⟶ onto dom ⁡ dom ⁡ G ↔ G : X × X ⟶ onto X
11 9 10 bitrd ⊢ dom ⁡ dom ⁡ G = X → G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G ↔ G : X × X ⟶ onto X
12 11 biimpd ⊢ dom ⁡ dom ⁡ G = X → G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G → G : X × X ⟶ onto X
13 5 12 mpcom ⊢ G : dom ⁡ dom ⁡ G × dom ⁡ dom ⁡ G ⟶ onto dom ⁡ dom ⁡ G → G : X × X ⟶ onto X
14 3 13 syl ⊢ G ∈ Magma ∩ ExId → G : X × X ⟶ onto X