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Proof step 4 from proof prove-ext (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
A.cols = B.cols, A.rows = B.rows, n < B.cols, m < B.rows, ∀ m, n. m < B.rows ∧ n < B.cols → A[m, n] = B[m, n] ⊦ B.cols = 0, B.rows = 0, A[m, n] = B[m, n]

Rule argument

Additional information

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