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Proof step 12 from proof append-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ , 1 + # x1 = # x0, ∀ x0, nf0, nf. # x1 = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf ⊦ a ' + x1 = x0 ∧ nf0 = nf

Rule argument

Additional information

Used simplifier rules: append-finfin: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ;% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)