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Proof step 5 from proof count-mkbag-greater-two (previous step)

A[m1] = A[m], A[m0] = A[m], m0 < # A, m < # A, m1 < # A, m0 < i, m < i, m1 < i, count(A[m], mkbag(A, i)) ≤ 2, i ≤ # A ⊦ m = m1, m = m0, m0 = m1, 2 ≤ count(A[m], mkbag(A, i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
A[m0] = A[m], A[m1] = A[m], m1 < i, m < i, m0 < i, m1 < # A, m < # A, m0 < # A, count(A[m], mkbag(A, i)) < 2, i ≤ # A ⊦ count(A[m], mkbag(A, i)) = 2, m = m0, m = m1, m0 = m1

Rule argument

Additional information

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