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Proof step 5 from proof count-mkbag-greater-one (previous step)

A[n] = A[m], n < i, m < i, n < # A, m < # A, i ≤ # A, 2 ≤ # A ⊦ m = n, count(A[m], mkbag(A, i)) ≠ 0 ∧ count(A[m], mkbag(A, i)) ≠ 1

User applied rule case distinction

And obtained the following new goals

Proof step 6:
A[n] = A[m], count(A[m], mkbag(A, i)) = 0, m < # A, n < # A, m < i, n < i, 2 ≤ # A, i ≤ # A ⊦ m = n
Proof step 8:
A[n] = A[m], count(A[m], mkbag(A, i)) = 1, m < # A, n < # A, m < i, n < i, 2 ≤ # A, i ≤ # A ⊦ m = n

Rule argument

pos: right 2

Additional information