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Proof step 31 from proof in-mkbag-del-two (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 32:
n < i, m < i, i ≤ # A, m ≠ n, A[m] = a, A[n] = a ⊦ count(a, mkbag(A, i)) > 1

Rule argument

spec: <toplevel> name: count-mkbag-greater-one seq: n < i, m < i, i ≤ # A, m ≠ n, A[m] = a, A[n] = a ⊦ count(a, mkbag(A, i)) > 1 subst: [n, i, m, A, a] → [m, i, n0, A, A[n0]]

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: m, i, n0, A, A[n0]