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Proof step 9 from proof count-mkbag-greater-one (previous step)

A[n] = A[m], count(A[m], mkbag(A, i0 +1)) = 1, n < i0 +1, m < i0 +1, n < # A, m < # A, i0 +1 ≤ # A, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
A[n] = A[m], count(A[m], mkbag(A, i0) ++ A[i0]) = 1, 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

Rule argument

Additional information

Used simplifier rules: mkbag-rec: ⊦ mkbag(A, n + 1) = mkbag(A, n) ++ A[n];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)