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Proof step 16 from proof diffgetn-nodupssize (previous step)

m0 < # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ 1 + #(x0 \ x0[m0] ') = # x0, a = x0[m0], x0 = [], dups(x0), a ∈ x0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: # x0 - 1, m0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s) Used terms: # x0 - 1, m0