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Proof step 2 from proof inapp-05 (previous step)

a ' + x + b ' + y = x0 + b ' + y0, a ∈ x0 ⊦ b ∈ x, b ∈ x0, a ' + x = x0 ∧ y = y0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x0

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x;%inapp-01: ⊦ ¬ a ∈ x ∧ ¬ a ∈ z → (x + a ' + y = z + a ' + y0 ↔ y = y0 ∧ x = z);%list freely generated by [], +;;