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Proof step 26 from proof count-mkbag-greater-two (previous step)

1 = count(A[i0], mkbag(A, i0)), A[m1] = A[i0], A[m0] = A[i0], m0 < i0, m1 < i0, m0 < # A, m1 < # A, i0 < # A, m0 + 1 < # A, m1 + 1 < # A ⊦ i0 = m1, i0 = m0, m0 = m1

User applied rule apply lemma

And obtained the following new goals

Proof step 27:
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] → [m1, i0, m0, A, A[i0]]

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m < n → m ≤ n; (s) Used terms: m1, i0, m0, A, A[i0]