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Proof step 14 from proof append-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# 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

Rule argument

Additional information

Used simplifier rules: append-finfin: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)