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

x + a ' + y0 = b0 ' + z0, x + a ' + y = b ' + z, b0 ∈ x ∨ b0 = a ∨ b0 ∈ y0, b ∈ x ∨ b = a ∨ b ∈ y ⊦ b = b0

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-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 [], +;;