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Proof step 56 from proof path-equiv-reach (previous step)

# x = 2 + n0, path(x), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ c. reach(x .first, c, n0) ∧ x .last ∈ succ(c)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 55:
x = a0 + y ∧ x ≠ [], # x = 2 + n0, path(x), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ c. reach(a0, c, n0) ∧ x .last ∈ succ(c)

Rule argument

[x0lemmaarg: spec-name: list-data instance-name: lemma-name: elim-list-c antecedent: x ≠ [] succedent: a = x .first ∧ y = x .rest ↔ x = a + y substlist: [x, a, y] → [x, a0, y] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0], [CSIMPSEQ : ⊦ m + n = 0 ↔ m = 0 ∧ n = 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPELIM : x ≠ [] ⊦ a = x .first ∧ y = x .rest ↔ x = a + y])], [EXTRATERMS : (x, a0, y)]]