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Proof step 16 from proof append-split (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ a ' + x = b ' + y ↔ a = b ∧ x = y
Proof step 18:
# x1 = # x, (⌜a '⌝ + ⌜x1⌝ ) + ⌜nf0⌝ = (⌜a0 '⌝ + ⌜x⌝ ) + ⌜nf⌝ , ∀ x0, nf0, nf. # x = # x0 ∧ ⌜x1⌝ + ⌜nf0⌝ = ⌜x0⌝ + ⌜nf⌝ → x1 = x0 ∧ nf0 = nf ⊦ (a = a0 ∧ x1 = x) ∧ nf0 = nf

Rule argument

spec: list name: c-06 seq: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y subst: [y, b, x, a] → [x, a0, x1, a] flags: <all paths>

Additional information

Used terms: x, a0, x1, a