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Proof step 4 from proof count-mkbag-non-zero (previous step)

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) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
count(A[n], mkbag(A, i0) ++ A[i0]) = 0, n < # A, i0 < # A, n ≤ i0, ∀ A0, n0. count(A0[n0], mkbag(A0, i0)) ≠ 0 ∨ ¬ n0 < i0 ∨ ¬ n0 < # A0 ∨ ¬ i0 ≤ # A0 ⊦

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)