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Proof step 2 from proof isin (previous step)

a ' + x = y + b ' + z, a ∈ y ⊦ y = [], b ∈ x

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: y

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z;%list freely generated by [], +;;