Metamath Proof Explorer


Theorem mpst123

Description: Decompose a pre-statement into a triple of values. (Contributed by Mario Carneiro, 18-Jul-2016)

Ref Expression
Hypothesis mpstssv.p ⊢ P = mPreSt ⁡ T
Assertion mpst123 ⊢ X ∈ P → X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X

Proof

Step Hyp Ref Expression
1 mpstssv.p ⊢ P = mPreSt ⁡ T
2 1 mpstssv ⊢ P ⊆ V × V × V
3 2 sseli ⊢ X ∈ P → X ∈ V × V × V
4 1st2nd2 ⊢ X ∈ V × V × V → X = 1 st ⁡ X 2 nd ⁡ X
5 xp1st ⊢ X ∈ V × V × V → 1 st ⁡ X ∈ V × V
6 1st2nd2 ⊢ 1 st ⁡ X ∈ V × V → 1 st ⁡ X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X
7 5 6 syl ⊢ X ∈ V × V × V → 1 st ⁡ X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X
8 7 opeq1d ⊢ X ∈ V × V × V → 1 st ⁡ X 2 nd ⁡ X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X
9 4 8 eqtrd ⊢ X ∈ V × V × V → X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X
10 df-ot ⊢ 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X
11 9 10 eqtr4di ⊢ X ∈ V × V × V → X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X
12 3 11 syl ⊢ X ∈ P → X = 1 st ⁡ 1 st ⁡ X 2 nd ⁡ 1 st ⁡ X 2 nd ⁡ X