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Proof step 3 from proof count-mkbag-non-zero (previous step)

count(A[n], mkbag(A, i)) = 0, n < i, n < # A, i ≤ # A ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
count(A[n], mkbag(A, i0 +1)) = 0, n < # A, n < i0 +1, i0 +1 ≤ # A, ∀ A, n. ¬(count(A[n], mkbag(A, i0)) = 0 ∧ n < i0 ∧ n < # A ∧ i0 ≤ # A) ⊦

Rule argument

indvar: i

Additional information

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