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Proof step 64 from proof INVR-step (previous step)

v ∈ C, reach(source, dest, n1 + n), reachS(source, C, n), reachS(source, N, n + 1), reach(b, dest, n1 - 1), b ∈ N, ∀ b. ¬ b ∈ N ∪ (succ(v) \ V) ∨ ¬ reach(b, dest, n1 - 1) ∨ reach<(b, dest, n1 - 1), ∀ a0. a0 = v ∨ ¬ a0 ∈ C ∨ ¬ reach(a0, dest, n1) ∨ reach<(a0, dest, n1) ⊦ N = ∅, # C = 0, v = dest, C = ∅, n1 = 0, # N = 0, reach<(source, dest, n1 + n), reach<(b, dest, n1 - 1), ∃ b0. ¬ (b0 ∈ V ↔ reach<(source, b0, n + 1) ∨ (∃ a0. b0 ∈ succ(a0) ∧ reach(source, a0, n) ∧ ¬ reach<(source, a0, n) ∧ ¬ a0 ∈ C)), ∃ b0. ¬ ((∃ c. reach(source, c, n) ∧ b0 ∈ succ(c)) ∧ ¬ reach<(source, b0, n + 1) → b0 ∈ N ∨ (∃ a. a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 63:
n1 = m + 1, v ∈ C, reach(source, dest, n1 + n), reachS(source, C, n), reachS(source, N, n + 1), reach(b, dest, m), b ∈ N, ∀ b. ¬ b ∈ N ∪ (succ(v) \ V) ∨ ¬ reach(b, dest, m) ∨ reach<(b, dest, m), ∀ a0. a0 = v ∨ ¬ a0 ∈ C ∨ ¬ reach(a0, dest, n1) ∨ reach<(a0, dest, n1) ⊦ N = ∅, # C = 0, v = dest, C = ∅, n1 = 0, # N = 0, reach<(source, dest, n1 + n), reach<(b, dest, m), ∃ b0. ¬ (b0 ∈ V ↔ reach<(source, b0, n + 1) ∨ (∃ a0. b0 ∈ succ(a0) ∧ reach(source, a0, n) ∧ ¬ reach<(source, a0, n) ∧ ¬ a0 ∈ C)), ∃ b0. ¬ ((∃ c. reach(source, c, n) ∧ b0 ∈ succ(c)) ∧ ¬ reach<(source, b0, n + 1) → b0 ∈ N ∨ (∃ a. a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V))

Rule argument

[x0lemmaarg: spec-name: nat instance-name: lemma-name: elim-pred-c antecedent: n ≠ 0 succedent: m = n - 1 ↔ n = m + 1 substlist: [n, m] → [n1, m] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n1, m)]]