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Proof step 5 from proof lastdiv (previous step)

x0 = y + (a0 + []), x0 = [] ∨ x0 = y + a0 ' ⊦ ∃ y0, a1. a ' + x0 = y0 + a1 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ ∃ y0, a1. a ' + y + a0 ' = y0 + a1 '

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%e-02: ⊦ x + a ' ≠ [];%app-01: ⊦ x + y = x + z ↔ y = z;