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

2 = count(A[m], mkbag(A, i0 +1)), A[m0] = A[m], A[m1] = A[m], m1 < i0 +1, m < i0 +1, m0 < i0 +1, m1 < # A, m < # A, m0 < # A, i0 +1 ≤ # A, ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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

Rule argument

Additional information

Used simplifier rules: mkbag-rec: ⊦ mkbag(A, n + 1) = mkbag(A, n) ++ A[n];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)