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 Description: Lemma 2 for smadiadet 20695: The summands of the Leibniz' formula vanish for all permutations fixing the index of the row containing the 0's and the 1 to itself. (Contributed by AV, 31-Dec-2018.)
Hypotheses
Ref Expression
marep01ma.a 𝐴 = (𝑁 Mat 𝑅)
marep01ma.b 𝐵 = (Base‘𝐴)
marep01ma.r 𝑅 ∈ CRing
marep01ma.0 0 = (0g𝑅)
marep01ma.1 1 = (1r𝑅)
Assertion
Ref Expression
smadiadetlem2 ((𝑀𝐵𝐾𝑁) → (𝑅 Σg (𝑝 ∈ (𝑃 ∖ {𝑞𝑃 ∣ (𝑞𝐾) = 𝐾}) ↦ (((𝑌𝑆)‘𝑝) · (𝐺 Σg (𝑛𝑁 ↦ (𝑛(𝑖𝑁, 𝑗𝑁 ↦ if(𝑖 = 𝐾, if(𝑗 = 𝐾, 1 , 0 ), (𝑖𝑀𝑗)))(𝑝𝑛))))))) = 0 )
Distinct variable groups:   𝑖,𝑗,𝑛,𝐵   𝑖,𝑞,𝐾,𝑗,𝑛   𝑖,𝑀,𝑗,𝑛   𝑖,𝑁,𝑗,𝑛   𝑃,𝑖,𝑗,𝑛,𝑞   𝑅,𝑖,𝑗,𝑛   1 ,𝑖,𝑗,𝑛   0 ,𝑖,𝑗,𝑛   𝑛,𝐺   𝑛,𝑝,𝐵   𝐾,𝑝   𝑀,𝑝   𝑁,𝑝   𝑃,𝑝   𝑅,𝑝,𝑖,𝑗   𝑞,𝑝
Allowed substitution hints:   𝐴(𝑖,𝑗,𝑛,𝑞,𝑝)   𝐵(𝑞)   𝑅(𝑞)   𝑆(𝑖,𝑗,𝑛,𝑞,𝑝)   · (𝑖,𝑗,𝑛,𝑞,𝑝)   1 (𝑞,𝑝)   𝐺(𝑖,𝑗,𝑞,𝑝)   𝑀(𝑞)   𝑁(𝑞)   𝑌(𝑖,𝑗,𝑛,𝑞,𝑝)   0 (𝑞,𝑝)

StepHypRef Expression
1 marep01ma.a . . 3 𝐴 = (𝑁 Mat 𝑅)
2 marep01ma.b . . 3 𝐵 = (Base‘𝐴)
3 marep01ma.r . . 3 𝑅 ∈ CRing
4 marep01ma.0 . . 3 0 = (0g𝑅)
5 marep01ma.1 . . 3 1 = (1r𝑅)