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Proof step 7 from proof count-mkbag-non-zero (previous step)

(A[n] = A[i0] ⊃ 1;0) + count(A[n], mkbag(A, i0)) = 0, i0 < # A, n < # A, n ≤ i0, ∀ A0, n0. count(A0[n0], mkbag(A0, i0)) ≠ 0 ∨ ¬ n0 < i0 ∨ ¬ n0 < # A0 ∨ ¬ i0 ≤ # A0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
count(A[n], mkbag(A, i0)) = 0, n < # A, i0 < # A, n ≤ i0, ∀ A0, n0. count(A0[n0], mkbag(A0, i0)) ≠ 0 ∨ ¬ n0 < i0 ∨ ¬ n0 < # A0 ∨ ¬ i0 ≤ # A0 ⊦ A[n] = A[i0]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)