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Proof step 22 from proof path-equiv-reach (previous step)

# x = n0 + 1, path(x), reach(x .first, x .last, n0), b ∈ succ(x .last) ⊦ # succ(x .last) = 0, succ(x .last) = ∅, ∃ x0. path(x0) ∧ # x0 = 2 + n0 ∧ x0 .first = x .first ∧ x0 .last = b

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
x = a + y ∧ x ≠ [], # x = n0 + 1, path(x), reach(a, x .last, n0), b ∈ succ(x .last) ⊦ # succ(x .last) = 0, succ(x .last) = ∅, ∃ x0. path(x0) ∧ # x0 = 2 + n0 ∧ x0 .first = a ∧ x0 .last = b

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, a, 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 + 1 ≠ 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, a, y)]]