Metamath Proof Explorer


Theorem satffunlem1

Description: Lemma 1 for satffun : induction basis. (Contributed by AV, 28-Oct-2023)

Ref Expression
Assertion satffunlem1 M V E W Fun M Sat E suc

Proof

Step Hyp Ref Expression
1 satfv0fun M V E W Fun M Sat E
2 satffunlem1lem1 Fun M Sat E Fun x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
3 1 2 syl M V E W Fun x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
4 satffunlem1lem2 M V E W dom M Sat E dom x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u =
5 funun Fun M Sat E Fun x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u dom M Sat E dom x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u = Fun M Sat E x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
6 1 3 4 5 syl21anc M V E W Fun M Sat E x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
7 peano1 ω
8 eqid M Sat E = M Sat E
9 8 satfvsuc M V E W ω M Sat E suc = M Sat E x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
10 7 9 mp3an3 M V E W M Sat E suc = M Sat E x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
11 10 funeqd M V E W Fun M Sat E suc Fun M Sat E x y | u M Sat E v M Sat E x = 1 st u 𝑔 1 st v y = M ω 2 nd u 2 nd v i ω x = 𝑔 i 1 st u y = f M ω | j M i j f ω i 2 nd u
12 6 11 mpbird M V E W Fun M Sat E suc