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Proof step 3 from proof index-in-mkbag (previous step)

m < # A, m < n0 +1, ∀ m, A. m < n0 ∧ m < # A → A[m] ∈ mkbag(A, n0) ⊦ A[m] ∈ mkbag(A, n0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m < # A, m ≤ n0, ∀ m, A. m < n0 ∧ m < # A → A[m] ∈ mkbag(A, n0) ⊦ A[m] = A[n0], A[m] ∈ mkbag(A, n0)

Rule argument

Additional information

Used simplifier rules: mkbag-rec: ⊦ mkbag(A, n + 1) = mkbag(A, n) ++ A[n];% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ a ∈ bag ++ b ↔ a ∈ bag ∨ a = b; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)