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Proof step 19 from proof append-split (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ (σ + σ1) + σ2 = σ + σ1 + σ2
Proof step 21:
# x1 = # x, ⌜a '⌝ + ⌜x1⌝ + ⌜nf0⌝ = (⌜a0 '⌝ + ⌜x⌝ ) + ⌜nf⌝ , ∀ x0, nf0, nf. # x = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf ⊦ a = a0 ∧ x1 = x ∧ nf0 = nf

Rule argument

spec: <toplevel> name: append-assoc seq: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2 subst: [σ2, σ1, σ] → [⌜nf0⌝ , ⌜x1⌝ , ⌜a '⌝ ] flags: <all paths>

Additional information

Used terms: ⌜nf0⌝ , ⌜x1⌝ , ⌜a '⌝