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Proof step 10 from proof count-mkbag-greater-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ count(a, bag ++ b) = (a = b ⊃ 1;0) + count(a, bag)
Proof step 12:
(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

Rule argument

spec: bag-basic name: count-insert seq: ⊦ count(a, bag ++ b) = (a = b ⊃ 1;0) + count(a, bag) subst: [b, bag, a] → [A[i0], mkbag(A, i0), A[m]] flags: <all paths>

Additional information

Used terms: A[i0], mkbag(A, i0), A[m]