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Proof step 5 from proof append-size-less (previous step)

x1 = a0 + x, (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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x3 + x0 = x + x2, # x3 < # x, ∀ x0, x1, x2. x3 + x0 = x1 + x2 ∧ # x3 < # x1 → x1 = [] ∨ (∃ y. x1 = x3 + y ∧ x0 = y + x2) ⊦ x = [], ∃ y. x = x3 + y ∧ x0 = y + x2

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%e-01: ⊦ a ' + x ≠ [];%len-05: ⊦ #(a ' + x) = (# x)+1;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2