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Proof step 13 from proof append-split (previous step)

(⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ , 1 + # x1 = # x0, ∀ x0, nf0, nf. # x1 = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf ⊦ a ' + x1 = x0 ∧ nf0 = nf

User applied rule structural induction

And obtained the following new goals

Proof step 14:
1 + # x1 = #(a0 + x), (⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = ⌜a0 + x⌝ + ⌜nf⌝ , ∀ x0, nf0, nf. # x1 = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf, ∀ a, x1, nf0, nf. (⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = ⌜x⌝ + ⌜nf⌝ ∧ 1 + # x1 = # x ∧ (∀ x0, nf0, nf. # x1 = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf) → a ' + x1 = x ∧ nf0 = nf ⊦ a ' + x1 = a0 + x ∧ nf0 = nf

Rule argument

indvar: x0

Additional information

Used simplifier rules: empty: ⊦ ⌜[]⌝ = [];%empty-append: ⊦ [] + σ = σ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic