[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 18 from proof correctness (previous step)

C3.cols = B.cols, C3.rows = A.rows, C4.cols = B.cols, C4.rows = A.rows, C2.cols = B.cols, C2.rows = A.rows, A.cols = B.rows, n + 1 < B.cols, n1 < B.cols, k < B.rows, m < A.rows, n < n1, n < B.cols, ∀ m0, n. m0 < A.rows ∧ n < B.cols → C3[m0, n] = C4[m0, n] + (m0 = m ∧ n < n1 ⊃ A[m, k] * B[k, n];0), ∀ m0, n. m0 < A.rows ∧ n < B.cols → C4[m0, n] = C2[m0, n] + (m0 = m ⊃ Σ(k, λ k. A[m, k] * B[k, n]);0), ∀ m1, n0. m1 < A.rows ∧ n0 < B.cols → C2[m1, n0] = (m1 < m ⊃ Σ(B.rows, λ k. A[m1, k] * B[k, n0]);0) ⊦ C3[m, n] = C4[m, n] + A[m, k] * B[k, n], B.cols = 0, A.rows = 0, B.rows = 0, n = n1, n1 = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 14 terms: m, n

Additional information

Used terms: m, n