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Proof step 2 from proof index-in-mkbag (previous step)

m < n, m < # A ⊦ A[m] ∈ mkbag(A, n)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
m < # A, m < n0 +1, ∀ m, A. m < n0 ∧ m < # A → A[m] ∈ mkbag(A, n0) ⊦ A[m] ∈ mkbag(A, n0 +1)

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ ¬ n < 0; (s)