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Proof step 25 from proof append-split (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: nf0, x, nf abort: F discard: T computed: T

Additional information

Used terms: nf0, x, nf