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Proof step 3 from proof count-mkbag-greater-one (previous step)

A[n] = A[m], n < i, m < i, n < # A, m < # A, i ≤ # A ⊦ m = n, count(A[m], mkbag(A, i)) ≠ 0 ∧ count(A[m], mkbag(A, i)) ≠ 1

User applied rule cut formula

And obtained the following new goals

Proof step 4:
A[n] = A[m], m < # A, n < # A, m < i, n < i, i ≤ # A, # A ≥ 2 ⊦ m = n, count(A[m], mkbag(A, i)) ≠ 0 ∧ count(A[m], mkbag(A, i)) ≠ 1

Rule argument

fma: # A ≥ 2

Additional information

Used simplifier rules: ⊦ n ≥ m ↔ m ≤ n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m +1 = m + 1; (s)%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s)% ⊦ m < 1 ↔ m = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: # A ≥ 2