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Proof step 15 from proof append-split (previous step)

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

User applied rule weakening

And obtained the following new goals

Rule argument

poslist: left 3

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