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Proof step 18 from proof last-elim (previous step)

a1 ' + x = y1 + a ', y1 = a0 + y, ∀ a2, y0. a2 = (y1 + a ').last ∧ y0 = (y1 + a ').butlast ↔ y1 + a ' = y0 + a2 ', a = a1 ∨ a ∈ x, a1 ∈ y1 ∨ a1 = a ⊦ a = (y1 + a ').last ∧ y1 = (y1 + a ').butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ a2, y0. a2 = ((a0 ' + y) + a ').last ∧ y0 = ((a0 ' + y) + a ').butlast ↔ (a0 ' + y) + a ' = y0 + a2 ' ⊦ a = ((a0 ' + y) + a ').last ∧ a0 ' + y = ((a0 ' + y) + a ').butlast

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list% ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)