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Proof step 5 from proof diffgetn-nodupssize (previous step)

∀ m, n. # x0 = m + 1 ∧ n ≤ m → #(x0 \ x0[n] ') = m ∨ x0 = [] ∨ dups(x0), #(a + x0) = m + 1, n ≤ m ⊦ #((a + x0) \ (a + x0)[n] ') = m, a + x0 = [], dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # x0, ∀ m0, n0. # x0 = m0 + 1 ∧ n0 ≤ m0 → #(x0 \ x0[n0] ') = m0 ∨ x0 = [] ⊦ #((a ' + x0) \ (a ' + x0)[n] ') = # x0, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list