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Proof step 5 from proof postfix-split-stronger (previous step)

x ⊒ y + z, ∀ x0, y, z. # x0 < # x → x0 ⊒ y + z → x0 ⊒ z ∨ (∃ x1. x1 + z = x0 ∧ x1 ⊒ y) ⊦ x ⊒ z, ∃ x0. x0 + z = x ∧ x0 ⊒ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] ∧ y ≠ [] → (x ⊒ y ↔ x.last = y.last ∧ x.butlast ⊒ y.butlast)
Proof step 7:
x.last = (y + z).last, x.butlast ⊒ (y + z).butlast, ∀ x0, y, z. # x0 < # x → x0 ⊒ y + z → x0 ⊒ z ∨ (∃ x1. x1 + z = x0 ∧ x1 ⊒ y) ⊦ y + z = [], x = [], x ⊒ z, ∃ x0. x0 + z = x ∧ x0 ⊒ y

Rule argument

spec: <toplevel> name: postfix-rec seq: ⊦ x ≠ [] ∧ y ≠ [] → (x ⊒ y ↔ x.last = y.last ∧ x.butlast ⊒ y.butlast) subst: [y, x] → [y + z, x] flags: <all paths>

Additional information

Used simplifier rules: postfix-empty: ⊦ [] ⊒ x;%postfix-empty-03: ⊦ x ⊒ [] ↔ x = [];% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s) Used terms: y + z, x