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Proof step 11 from proof count-mkbag-greater-two (previous step)

2 = count(A[m], mkbag(A, i0) ++ A[i0]), A[m1] = A[m], A[m0] = A[m], i0 < # A, m0 < # A, m < # A, m1 < # A, m0 ≤ i0, m ≤ i0, m1 ≤ i0, ∀ A, m, m1, m0. 2 = count(A[m], mkbag(A, i0)) ∧ A[m1] = A[m] ∧ A[m0] = A[m] ∧ m0 < # A ∧ m < # A ∧ m1 < # A ∧ m0 < i0 ∧ m < i0 ∧ m1 < i0 ∧ i0 ≤ # A → m = m1 ∨ m = m0 ∨ m0 = m1 ⊦ m = m1, m = m0, m0 = m1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ count(a, bag ++ b) = (a = b ⊃ 1;0) + count(a, bag)
Proof step 13:
2 = (A[m] = A[i0] ⊃ 1;0) + count(A[m], mkbag(A, i0)), A[m0] = A[m], A[m1] = A[m], m1 < # A, m < # A, m0 < # A, i0 < # A, m1 ≤ i0, m ≤ i0, m0 ≤ i0, ∀ A, m, m1, m0. 2 = count(A[m], mkbag(A, i0)) ∧ A[m1] = A[m] ∧ A[m0] = A[m] ∧ m0 < # A ∧ m < # A ∧ m1 < # A ∧ m0 < i0 ∧ m < i0 ∧ m1 < i0 ∧ i0 ≤ # A → m = m1 ∨ m = m0 ∨ m0 = m1 ⊦ m = m0, m = m1, m0 = m1

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]