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Proof step 6 from proof in-mkbag-del-two (previous step)

A[m] = A[n0], count(A[m], mkbag(A, i) -- A[n] -- A[n0]) = 0, m < i, n < i, n0 < i, m < # A, n < # A, n0 < # A, i ≤ # A ⊦ n = n0, n = m, n0 = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
count(A[n0], mkbag(A, i) -- A[n]) - (count(A[n0], mkbag(A, i) -- A[n]) = 0 ⊃ 0;1) = 0, A[m] = A[n0], m < i, n < i, n0 < i, m < # A, n < # A, n0 < # A, i ≤ # A ⊦ n = n0, n = m, n0 = m

Rule argument

Additional information

Used simplifier rules: count-del-same: ⊦ count(a, bag -- a) = count(a, bag) - (count(a, bag) = 0 ⊃ 0;1);