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Proof step 16 from proof count-mkbag-greater-one (previous step)

A[m] = A[i0], A[n] = A[i0], count(A[i0], mkbag(A, i0)) = 0, i0 = m, n < # A, m < # A, i0 < # A, n ≤ i0, m ≤ i0, 2 ≤ # A ⊦ m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
A[n] = A[m], count(A[m], mkbag(A, m)) = 0, n < m, n < # A, m < # A, n + 1 < # A, 2 ≤ # A ⊦ m = n

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ m; (s)