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Proof step 4 from proof diffgetn-nodupssize (previous step)

# x = m + 1, n < # x ⊦ #(x \ x[n] ') = m, # x = 0, x = [], dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x = m + 1, n ≤ m ⊦ #(x \ x[n] ') = m, x = [], dups(x)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)