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Proof step 6 from proof diffgetn-nodupssize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ a ' + x ≠ []; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)