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Proof step 7 from proof in-mkbag-del (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
count(A[n], mkbag(A, i)) - 0 = 0, A[m] = A[n], count(A[n], mkbag(A, i)) = 0, n < # A, m < # A, n < i, m < i, i ≤ # A ⊦ n = m
Proof step 11:
count(A[n], mkbag(A, i)) - 1 = 0, A[m] = A[n], n < # A, m < # A, n < i, m < i, i ≤ # A ⊦ count(A[n], mkbag(A, i)) = 0, n = m

Rule argument

pos: left 1

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