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Proof step 2 from proof two (previous step)

a ' + b ' = x + a0 ' + c ' + y, a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ y ⊦ x = [] ∧ y = [] ∧ a = a0 ∧ b = c

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, b, x, a0, c. a ' + b ' = x + a0 ' + c ' + y0 ∧ (a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ y0) → x = [] ∧ y0 = [] ∧ a = a0 ∧ b = c, a ' + b ' = x + a0 ' + c ' + (a1 + y0), a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ a1 + y0 ⊦ x = [] ∧ a1 + y0 = [] ∧ a = a0 ∧ b = c

Rule argument

term: y

Additional information

Used simplifier rules: app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b;%c-05: ⊦ a ' = x + b ' ↔ a = b ∧ x = [];%list freely generated by [], +;;