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Proof step 22 from proof in-mkbag-del-two (previous step)

A[m] = A[n], A[n0] = A[n], count(A[n], mkbag(A, i)) = 2, n0 < # A, n < # A, m < # A, n0 < i, n < i, m < i, i ≤ # A ⊦ n = m, n = n0, n0 = m

User applied rule apply lemma

And obtained the following new goals

Proof step 23:
m1 < i, m < i, m0 < i, i ≤ # A, m ≠ m1, m0 ≠ m1, m ≠ m0, A[m] = a, A[m0] = a, A[m1] = a ⊦ count(a, mkbag(A, i)) > 2

Rule argument

spec: <toplevel> name: count-mkbag-greater-two seq: m1 < i, m < i, m0 < i, i ≤ # A, m ≠ m1, m0 ≠ m1, m ≠ m0, A[m] = a, A[m0] = a, A[m1] = a ⊦ count(a, mkbag(A, i)) > 2 subst: [m1, i, m, m0, A, a] → [n, i, n0, m, A, A[n]]

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s) Used terms: n, i, n0, m, A, A[n]