Metamath Proof Explorer


Theorem ismfs

Description: A formal system is a tuple <. mCN , mVR , mType , mVT , mTC , mAx >. such that: mCN and mVR are disjoint; mType is a function from mVR to mVT ; mVT is a subset of mTC ; mAx is a set of statements; and for each variable typecode, there are infinitely many variables of that type. (Contributed by Mario Carneiro, 18-Jul-2016)

Ref Expression
Hypotheses ismfs.c C = mCN T
ismfs.v V = mVR T
ismfs.y Y = mType T
ismfs.f F = mVT T
ismfs.k K = mTC T
ismfs.a A = mAx T
ismfs.s S = mStat T
Assertion ismfs T W T mFS C V = Y : V K A S v F ¬ Y -1 v Fin

Proof

Step Hyp Ref Expression
1 ismfs.c C = mCN T
2 ismfs.v V = mVR T
3 ismfs.y Y = mType T
4 ismfs.f F = mVT T
5 ismfs.k K = mTC T
6 ismfs.a A = mAx T
7 ismfs.s S = mStat T
8 fveq2 t = T mCN t = mCN T
9 8 1 eqtr4di t = T mCN t = C
10 fveq2 t = T mVR t = mVR T
11 10 2 eqtr4di t = T mVR t = V
12 9 11 ineq12d t = T mCN t mVR t = C V
13 12 eqeq1d t = T mCN t mVR t = C V =
14 fveq2 t = T mType t = mType T
15 14 3 eqtr4di t = T mType t = Y
16 fveq2 t = T mTC t = mTC T
17 16 5 eqtr4di t = T mTC t = K
18 15 11 17 feq123d t = T mType t : mVR t mTC t Y : V K
19 13 18 anbi12d t = T mCN t mVR t = mType t : mVR t mTC t C V = Y : V K
20 fveq2 t = T mAx t = mAx T
21 20 6 eqtr4di t = T mAx t = A
22 fveq2 t = T mStat t = mStat T
23 22 7 eqtr4di t = T mStat t = S
24 21 23 sseq12d t = T mAx t mStat t A S
25 fveq2 t = T mVT t = mVT T
26 25 4 eqtr4di t = T mVT t = F
27 15 cnveqd t = T mType t -1 = Y -1
28 27 imaeq1d t = T mType t -1 v = Y -1 v
29 28 eleq1d t = T mType t -1 v Fin Y -1 v Fin
30 29 notbid t = T ¬ mType t -1 v Fin ¬ Y -1 v Fin
31 26 30 raleqbidv t = T v mVT t ¬ mType t -1 v Fin v F ¬ Y -1 v Fin
32 24 31 anbi12d t = T mAx t mStat t v mVT t ¬ mType t -1 v Fin A S v F ¬ Y -1 v Fin
33 19 32 anbi12d t = T mCN t mVR t = mType t : mVR t mTC t mAx t mStat t v mVT t ¬ mType t -1 v Fin C V = Y : V K A S v F ¬ Y -1 v Fin
34 df-mfs mFS = t | mCN t mVR t = mType t : mVR t mTC t mAx t mStat t v mVT t ¬ mType t -1 v Fin
35 33 34 elab2g T W T mFS C V = Y : V K A S v F ¬ Y -1 v Fin