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Proof step 2 from proof count-mkbag-greater-one (previous step)

A[n] = a, A[m] = a, m < i, n < i, i ≤ # A ⊦ m = n, count(a, mkbag(A, i)) > 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
A[n] = A[m], n < i, m < i, n < # A, m < # A, i ≤ # A ⊦ m = n, count(A[m], mkbag(A, i)) ≠ 0 ∧ count(A[m], mkbag(A, i)) ≠ 1

Rule argument

Additional information

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