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Proof step 2 from proof dups-rmdup-size-equiv (previous step)

# x = # rmdup(x), dups(x) ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: rmdup-e: ⊦ rmdup([]) = [];%e-01: ⊦ ¬ dups([]);%rec: ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x);%rmdup-y: ⊦ a ∈ x → rmdup(a ' + x) = rmdup(x);%rm-size-not: ⊦ # rmdup(x) ≠ # x + 1;%rmdup-n: ⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x);%list freely generated by [], +;;% ⊦ a + x = a ' + x; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # [] = 0; (s)