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Proof step 10 from proof correctness (previous step)

A.cols = B.rows, m2 = 0, C1 = A.rows × B.cols, m ≤ C2.rows, C2.rows = A.rows, C2.cols = B.cols, ∀ m0, n0. m0 < C2.rows ∧ n0 < C2.cols → C2[m0, n0] = C1[m0, n0] + (m0 < m ∧ n0 < C2.cols ⊃ Σ(B.rows, λ k. A[m0, k] * B[k, n0]);0), C2 = C0, m = m1, m < C2.rows, k1 = 0, k ≤ B.rows, C4.rows = A.rows, C4.cols = B.cols, ∀ m0, n0. m0 < C4.rows ∧ n0 < C4.cols → C4[m0, n0] = C2[m0, n0] + (m0 = m ∧ n0 < C4.cols ⊃ Σ(k, λ k. A[m0, k] * B[k, n0]);0), k = k0, k < B.rows, n2 = 0, n ≤ C.cols, C.rows = A.rows, C.cols = B.cols, ∀ m0, n0. m0 < C.rows ∧ n0 < C.cols → C[m0, n0] = C4[m0, n0] + (m0 = m ∧ n0 < n ⊃ A[m0, k] * B[k, n0];0), C = C3, n = n1, n < C.cols ⊦ ⟨C[m, n] := C[m, n] + A[m, k] * B[k, n]⟩ ⟨n := n + 1⟩ ( ( n ≤ C.cols ∧ C.rows = A.rows ∧ C.cols = B.cols ∧ (∀ m0, n0. m0 < C.rows ∧ n0 < C.cols → C[m0, n0] = C4[m0, n0] + (m0 = m ∧ n0 < n ⊃ A[m0, k] * B[k, n0];0))) ∧ C.cols - n < C3.cols - n1)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 11:
A.cols = B.rows, m2 = 0, C1 = A.rows × B.cols, m ≤ C2.rows, C2.rows = A.rows, C2.cols = B.cols, ∀ m0, n0. m0 < C2.rows ∧ n0 < C2.cols → C2[m0, n0] = C1[m0, n0] + (m0 < m ∧ n0 < C2.cols ⊃ Σ(B.rows, λ k. A[m0, k] * B[k, n0]);0), C2 = C0, m = m1, m < C2.rows, k1 = 0, k ≤ B.rows, C4.rows = A.rows, C4.cols = B.cols, ∀ m0, n0. m0 < C4.rows ∧ n0 < C4.cols → C4[m0, n0] = C2[m0, n0] + (m0 = m ∧ n0 < C4.cols ⊃ Σ(k, λ k. A[m0, k] * B[k, n0]);0), k = k0, k < B.rows, n2 = 0, n ≤ C.cols, C.rows = A.rows, C.cols = B.cols, ∀ m0, n0. m0 < C.rows ∧ n0 < C.cols → C[m0, n0] = C4[m0, n0] + (m0 = m ∧ n0 < n ⊃ A[m0, k] * B[k, n0];0), C = C3, n = n1, n < C.cols ⊦ ⟨n0 := n + 1⟩ ( ( n0 ≤ C[m, n, C[m, n] + A[m, k] * B[k, n]].cols ∧ C[m, n, C[m, n] + A[m, k] * B[k, n]].rows = A.rows ∧ C[m, n, C[m, n] + A[m, k] * B[k, n]].cols = B.cols ∧ (∀ m0, n1. m0 < C[m, n, C[m, n] + A[m, k] * B[k, n]].rows ∧ n1 < C[m, n, C[m, n] + A[m, k] * B[k, n]].cols → C[m, n, C[m, n] + A[m, k] * B[k, n]][m0, n1] = C4[m0, n1] + (m0 = m ∧ n1 < n0 ⊃ A[m0, k] * B[k, n1];0))) ∧ C[m, n, C[m, n] + A[m, k] * B[k, n]].cols - n0 < C3.cols - n1)

Proof step 10 from proof correctness (previous step)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

Additional information