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Proof step 11 from proof append-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ , 1 + # x1 = # x0, ∀ nf0, x0, nf. ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ ∧ # x1 = # x0 → x1 = x0 ∧ nf0 = nf ⊦ a ' + x1 = x0 ∧ nf0 = nf

Rule argument

Additional information

Used simplifier rules: append-finfin: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ;%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list