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Proof step 5 from proof first-02 (previous step)

x + a ' = a1 ' + y1, x + a ' + y = a0 ' + y0, a0 ∈ x ∨ a0 = a ∨ a0 ∈ y, a = a1 ∨ a ∈ y1, a1 ∈ x ∨ a1 = a ⊦ a0 = a1

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-03: ⊦ a ' = b ' + x ↔ a = b ∧ x = [];%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;%list freely generated by [], +;;