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Proof step 4 from proof last-ex (previous step)

x = y0 + a0 ' ∧ x ≠ [], a ∈ x ⊦ a = a0, ∃ y, z. x = y + a ' + z ∧ z ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ y0 ⊦ a = a0, ∃ y, z. y0 + a0 ' = y + a ' + z ∧ z ≠ []

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-02: ⊦ x + a ' ≠ []; from specification list