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Proof step 13 from proof count-mkbag-greater-one (previous step)

1 + count(A[m], mkbag(A, i0)) = 1, A[n] = A[m], A[m] = A[i0], i0 < # A, m < # A, n < # A, 2 ≤ # A, m ≤ i0, n ≤ i0, ∀ A, n, m. A[n] = A[m] ∧ count(A[m], mkbag(A, i0)) = 1 ∧ m < # A ∧ n < # A ∧ m < i0 ∧ n < i0 ∧ 2 ≤ # A ∧ i0 ≤ # A → m = n ⊦ m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
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, ∀ A, n, m. A[n] = A[m] ∧ count(A[m], mkbag(A, i0)) = 1 ∧ m < # A ∧ n < # A ∧ m < i0 ∧ n < i0 ∧ 2 ≤ # A ∧ i0 ≤ # A → m = n ⊦ m = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 1 = m + 1 ↔ m = 0; (s)