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Proof step 12 from proof count-mkbag-greater-one (previous step)

(A[m] = A[i0] ⊃ 1;0) + count(A[m], mkbag(A, i0)) = 1, A[n] = A[m], 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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
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
Proof step 22:
0 + count(A[m], mkbag(A, i0)) = 1, A[n] = A[m], 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 ⊦ A[m] = A[i0], m = n

Rule argument

pos: left 1

Proof step 12 from proof count-mkbag-greater-one (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information