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Proof step 2 from proof mkbag-update (previous step)

m < n, m < # A ⊦ mkbag(A[m, a], n) = (mkbag(A, n) -- A[m]) ++ a

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: n

Additional information

Used simplifier rules: mkbag-base: ⊦ mkbag(A, 0) = ∅;%nat freely generated by 0, +1;;% ⊦ ∅ -- a = ∅; (s)% ⊦ bag ++ a ≠ ∅; (s)% ⊦ ¬ n < 0; (s)