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Proof step 6 from proof append-size-less (previous step)

(a ' + x3) + x0 = x1 + x2, x1 = a0 + x, 1 + # x3 < # 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 7:
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: ⊦ #(x + y) = # x + # y;%len-02: ⊦ # a ' = 0 +1;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%in-17: ⊦ a ∈ b ' ↔ a = b;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)