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Proof step 1 from proof perm-one (conclusion)

⊦ perm(x, a ') ↔ x = a '

User applied rule structural induction

And closed the goal!

Rule argument

indvar: x

Additional information

Used simplifier rules: empty-01: ⊦ perm([], y) ↔ y = [];%del-01: ⊦ perm(a ' + x, y) ↔ a ∈ y ∧ perm(x, y -1l a);%empty: ⊦ perm(x, []) ↔ x = [];%ref: ⊦ perm(x, x);%list freely generated by [], +;;% ⊦ a + x = a ' + x; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ' -1l a = []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)% ⊦ a ' = b ' ↔ a = b; (s)% ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws)% ⊦ a ' ≠ []; (s)