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Proof step 35 from proof path-equiv-reach (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
x .last = y .butlast .last, # x = n0 + 1, # y = 1 + n0, path(x), path(x .first ' + y), reach(x .first, x .last, n0) ⊦ y = [], ∃ c. reach(x .first, c, n0) ∧ y .last ∈ succ(c)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]