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Proof step 20 from proof append-split (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ (σ + σ1) + σ2 = σ + σ1 + σ2
Proof step 22:
⌜a '⌝ + ⌜x1⌝ + ⌜nf0⌝ = ⌜a0 '⌝ + ⌜x⌝ + ⌜nf⌝ , # x1 = # x, ∀ nf0, x0, nf. ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ ∧ # x = # x0 → x1 = x0 ∧ nf0 = nf ⊦ a = a0 ∧ x1 = x ∧ nf0 = nf

Rule argument

spec: stream inst: name: append-assoc seq: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2 subst: [σ2, σ1, σ] → [⌜nf⌝ , ⌜x⌝ , ⌜a0 '⌝ ] flags: rewrite, current

Additional information

Used terms: ⌜nf⌝ , ⌜x⌝ , ⌜a0 '⌝