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Proof step 49 from proof path-equiv-reach (previous step)

# y = 1 + n0, path(a0 ' + y), ∀ x. ¬ path(x) ∨ # x ≠ n0 + 1 ∨ x .first ≠ a0 ∨ x .last ≠ y .butlast .last ⊦ y = [], reach(a0, y .butlast .last, n0), ∃ c. reach(a0, c, n0) ∧ y .last ∈ succ(c)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 48:
y = y0 + a ', # y = 1 + n0, path(a0 ' + y), ∀ x. ¬ path(x) ∨ # x ≠ n0 + 1 ∨ x .first ≠ a0 ∨ x .last ≠ y0 .last ⊦ y = [], reach(a0, y0 .last, n0), ∃ c. reach(a0, c, n0) ∧ a ∈ succ(c)

Rule argument

[x0lemmaarg: spec-name: list-last instance-name: lemma-name: last-elim antecedent: x ≠ [] succedent: a = x .last ∧ y = x .butlast ↔ x = y + a ' substlist: [x, a, y] → [y, a, y0] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : x ≠ [] ⊦ a = x .last ∧ y = x .butlast ↔ x = y + a '])], [EXTRATERMS : (y, a, y0)]]