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Proof step 20 from proof path-equiv-reach (previous step)

b ∈ succ((a ' + y) .last), reach(a, (a ' + y) .last, # y), path(a ' + y) ⊦ succ((a ' + y) .last) = ∅, # succ((a ' + y) .last) = 0, ∃ x0. path(x0) ∧ # x0 = 2 + # y ∧ x0 .first = a ∧ x0 .last = b

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 14:
y = [], b ∈ succ((a ' + y) .last), reach(a, (a ' + y) .last, # y), path(a ' + y) ⊦ succ((a ' + y) .last) = ∅, # succ((a ' + y) .last) = 0, ∃ x0. path(x0) ∧ # x0 = 2 + # y ∧ x0 .first = a ∧ x0 .last = b
Proof step 19:
b ∈ succ((a ' + y) .last), reach(a, (a ' + y) .last, # y), path(a ' + y) ⊦ y = [], succ((a ' + y) .last) = ∅, # succ((a ' + y) .last) = 0, ∃ x0. path(x0) ∧ # x0 = 2 + # y ∧ x0 .first = a ∧ x0 .last = b

Rule argument

[fma-arg: y = []]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (y = [])]]