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Proof step 3 from proof count-mkbag-greater-two (previous step)

A[m0] = A[m], A[m1] = A[m], m1 < i, m < i, m0 < i, m1 < # A, m < # A, m0 < # A, i ≤ # A, count(A[m], mkbag(A, i)) ≤ 2 ⊦ m = m0, m = m1, m0 = m1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ m ≤ n → (n ≤ m ↔ m = n)
Proof step 5:
A[m1] = A[m], A[m0] = A[m], m0 < # A, m < # A, m1 < # A, m0 < i, m < i, m1 < i, count(A[m], mkbag(A, i)) ≤ 2, i ≤ # A ⊦ m = m1, m = m0, m0 = m1, 2 ≤ count(A[m], mkbag(A, i))
Proof step 8:
2 = count(A[m], mkbag(A, i)), A[m1] = A[m], A[m0] = A[m], m0 < # A, m < # A, m1 < # A, m0 < i, m < i, m1 < i, i ≤ # A, 2 ≤ count(A[m], mkbag(A, i)) ⊦ m = m1, m = m0, m0 = m1

Rule argument

spec: nat name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [2, count(A[m], mkbag(A, i))] flags: <all paths>

Additional information

Used terms: 2, count(A[m], mkbag(A, i))