Metamath Proof Explorer


Theorem setsn0fun

Description: The value of the structure replacement function (without the empty set) is a function if the structure (without the empty set) is a function. (Contributed by AV, 7-Jun-2021) (Revised by AV, 16-Nov-2021)

Ref Expression
Hypotheses setsn0fun.s ⊢ φ → S Struct X
setsn0fun.i ⊢ φ → I ∈ U
setsn0fun.e ⊢ φ → E ∈ W
Assertion setsn0fun ⊢ φ → Fun ⁡ S sSet I E ∖ ∅

Proof

Step Hyp Ref Expression
1 setsn0fun.s ⊢ φ → S Struct X
2 setsn0fun.i ⊢ φ → I ∈ U
3 setsn0fun.e ⊢ φ → E ∈ W
4 structn0fun ⊢ S Struct X → Fun ⁡ S ∖ ∅
5 structex ⊢ S Struct X → S ∈ V
6 setsfun0 ⊢ S ∈ V ∧ Fun ⁡ S ∖ ∅ ∧ I ∈ U ∧ E ∈ W → Fun ⁡ S sSet I E ∖ ∅
7 5 6 sylanl1 ⊢ S Struct X ∧ Fun ⁡ S ∖ ∅ ∧ I ∈ U ∧ E ∈ W → Fun ⁡ S sSet I E ∖ ∅
8 7 expcom ⊢ I ∈ U ∧ E ∈ W → S Struct X ∧ Fun ⁡ S ∖ ∅ → Fun ⁡ S sSet I E ∖ ∅
9 2 3 8 syl2anc ⊢ φ → S Struct X ∧ Fun ⁡ S ∖ ∅ → Fun ⁡ S sSet I E ∖ ∅
10 9 com12 ⊢ S Struct X ∧ Fun ⁡ S ∖ ∅ → φ → Fun ⁡ S sSet I E ∖ ∅
11 4 10 mpdan ⊢ S Struct X → φ → Fun ⁡ S sSet I E ∖ ∅
12 1 11 mpcom ⊢ φ → Fun ⁡ S sSet I E ∖ ∅