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Proof step 8 from proof count-mkbag-greater-one (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 9:
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

Rule argument

indvar: i

Additional information

Used simplifier rules: mkbag-base: ⊦ mkbag(A, 0) = ∅;%nat freely generated by 0, +1;;% ⊦ count(a, ∅) = 0; (s)