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Proof step 3 from proof append-size-less (previous step)

∀ x0, x1, x2. x3 + x0 = x1 + x2 ∧ # x3 < # x1 → x1 = [] ∨ (∃ y. x1 = x3 + y ∧ x0 = y + x2), (a + x3) + x0 = x1 + x2, #(a + x3) < # x1 ⊦ x1 = [], ∃ y. x1 = (a + x3) + y ∧ x0 = y + x2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a ' + x3) + x0 = x1 + x2, # x3 + 1 < # x1, ∀ x0, x1, x2. x3 + x0 = x1 + x2 ∧ # x3 < # x1 → x1 = [] ∨ (∃ y. x1 = x3 + y ∧ x0 = y + x2), a ∈ x1 ∨ a ∈ x2 ⊦ x1 = [], ∃ y. x1 = (a ' + x3) + y ∧ x0 = y + x2

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%len-05: ⊦ #(a ' + x) = (# x)+1;%in-14: ⊦ a ' + x = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%succ: ⊦ m +1 = m + 1; from specification nat-basic2