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Proof step 2 from proof inex-01 (previous step)

a ' + x = y + b ' + z, a ∈ y ⊦ a = b, ∃ y0. y = a ' + y0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: y

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);%list freely generated by [], +;;