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Proof step 24 from proof append-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⌜x1⌝ + ⌜nf0⌝ = ⌜x⌝ + ⌜nf⌝ , # x1 = # x, ∀ nf0, x0, nf. ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ ∧ # x = # x0 → x1 = x0 ∧ nf0 = nf ⊦ x1 = x ∧ nf0 = nf

Proof step 24 from proof append-split (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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