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Proof step 3 from proof in-mkbag-del-two (previous step)

m < i, n < i, n0 < i, m < # A, n < # A, n0 < # A, i ≤ # A ⊦ n = n0, n = m, n0 = m, A[m] ∈ mkbag(A, i) -- A[n] -- A[n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 5:
count(A[m], mkbag(A, i) -- A[n] -- A[n0]) = 0, n0 < # A, n < # A, m < # A, n0 < i, n < i, m < i, i ≤ # A ⊦ n = m, n = n0, n0 = m

Rule argument

spec: bag-union name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [mkbag(A, i) -- A[n] -- A[n0], A[m]] flags: <all paths>

Additional information

Used terms: mkbag(A, i) -- A[n] -- A[n0], A[m]