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Proof step 54 from proof path-equiv-reach (previous step)

# y = 1 + n0, path(a0 ' + y), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ c. reach(a0, c, n0) ∧ (a0 ' + y) .last ∈ succ(c)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 53:
# y = 1 + n0, path(a0 ' + y), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ y = [], ∃ c. reach(a0, c, n0) ∧ (a0 ' + y) .last ∈ succ(c)

Rule argument

[fma-arg: y = []]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRAFMAS : (y = [])]]