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Proof step 8 from proof in-mkbag-del-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
count(A[n], mkbag(A, i)) - (count(A[n], mkbag(A, i)) = 0 ⊃ 0;1) - (count(A[n], mkbag(A, i)) - (count(A[n], mkbag(A, i)) = 0 ⊃ 0;1) = 0 ⊃ 0;1) = 0, A[m] = A[n], A[n0] = A[n], n0 < # A, n < # A, m < # A, n0 < i, n < i, m < i, i ≤ # A ⊦ n = m, n = n0, 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);