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Proof step 61 from proof correctness (previous step)

m1 < m, A.cols = B.rows, C0.rows = A.rows, C0.cols = B.cols, C.rows = A.rows, C.cols = B.cols, n < B.cols, m < A.rows, m1 < A.rows, ∀ m0, n0. m0 < A.rows ∧ n0 < B.cols → C0[m0, n0] = (m0 < m1 ⊃ Σ(B.rows, λ k. A[m0, k] * B[k, n0]);0), ∀ m, n. m < A.rows ∧ n < B.cols → C[m, n] = C0[m, n] + (m = m1 ⊃ Σ(B.rows, λ k. A[m1, k] * B[k, n]);0) ⊦ C[m, n] = 0, B.cols = 0, A.rows = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
A.cols = B.rows, C0.rows = A.rows, C0.cols = B.cols, C.rows = A.rows, C.cols = B.cols, m1 + 1 < A.rows, m1 < m, n < B.cols, m < A.rows, m1 < A.rows, ∀ m0, n0. m0 < A.rows ∧ n0 < B.cols → C0[m0, n0] = (m0 < m1 ⊃ Σ(B.rows, λ k. A[m0, k] * B[k, n0]);0), ∀ m, n. m < A.rows ∧ n < B.cols → C[m, n] = C0[m, n] + (m = m1 ⊃ Σ(B.rows, λ k. A[m1, k] * B[k, n]);0) ⊦ C[m, n] = 0, B.cols = 0, A.rows = 0, m = 0

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1