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Proof step 15 from proof count-mkbag-greater-one (previous step)

A[n] = A[i0], A[m] = A[i0], count(A[i0], mkbag(A, i0)) = 0, i0 < # A, m < # A, n < # A, 2 ≤ # A, m ≤ i0, n ≤ i0 ⊦ m = n

User applied rule cut formula

And obtained the following new goals

Proof step 16:
A[m] = A[i0], A[n] = A[i0], count(A[i0], mkbag(A, i0)) = 0, i0 = m, n < # A, m < # A, i0 < # A, n ≤ i0, m ≤ i0, 2 ≤ # A ⊦ m = n
Proof step 19:
A[m] = A[i0], A[n] = A[i0], count(A[i0], mkbag(A, i0)) = 0, n < # A, m < # A, i0 < # A, n ≤ i0, m ≤ i0, 2 ≤ # A ⊦ i0 = m, m = n

Rule argument

fma: i0 = m

Additional information

Used formulas: i0 = m