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Proof step 11 from proof append-split (previous step)

# x = # x0, ⌜x⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ ⊦ x = x0 ∧ nf0 = nf

User applied rule structural induction

And obtained the following new goals

Proof step 12:
#(a + x1) = # x0, ⌜a + x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ , ∀ x0, nf0, nf. # x1 = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf ⊦ a + x1 = x0 ∧ nf0 = nf

Rule argument

indvar: x

Additional information

Used simplifier rules: empty: ⊦ ⌜[]⌝ = [];%empty-append: ⊦ [] + σ = σ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)