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Proof step 5 from proof dot-assoc (previous step)

A.cols = B.rows, B.cols = C.rows ⊦ ((A * B) * C).rows = (A * B * C).rows ∧ ((A * B) * C).cols = (A * B * C).cols ∧ (m < ((A * B) * C).rows ∧ n < ((A * B) * C).cols → ((A * B) * C)[m, n] = (A * B * C)[m, n])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ A.cols = B.rows → (A * B).rows = A.rows
Proof step 7:
(A * B).cols = C.rows, A.cols = B.rows, B.cols = C.rows ⊦ (A * B).rows = (A * B * C).rows ∧ ((A * B) * C).cols = (A * B * C).cols ∧ (m < (A * B).rows ∧ n < ((A * B) * C).cols → ((A * B) * C)[m, n] = (A * B * C)[m, n])

Rule argument

spec: inst: name: rows-dot seq: ⊦ A.cols = B.rows → (A * B).rows = A.rows subst: [B, A] → [C, A * B] flags: rewrite, current

Additional information

Used simplifier rules: cols-dot: ⊦ A.cols = B.rows → (A * B).cols = B.cols; Used terms: C, A * B