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Proof step 5 from proof in-mkbag-del-two (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
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
Proof step 33:
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 ⊦ A[m] = A[n0], n = n0, n = m, n0 = m

Rule argument

fma: A[m] = A[n0]

Additional information

Used formulas: A[m] = A[n0]