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Proof step 48 from proof path-equiv-reach (previous step)

y = y0 + a ', # y = 1 + n0, path(a0 ' + y), ∀ x. ¬ path(x) ∨ # x ≠ n0 + 1 ∨ x .first ≠ a0 ∨ x .last ≠ y0 .last ⊦ y = [], reach(a0, y0 .last, n0), ∃ c. reach(a0, c, n0) ∧ a ∈ succ(c)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
path(a0 ' + y0 + a '), ∀ x. ¬ path(x) ∨ # x ≠ # y0 + 1 ∨ x .first ≠ a0 ∨ x .last ≠ y0 .last ⊦ reach(a0, y0 .last, # y0), ∃ c. reach(a0, c, # y0) ∧ a ∈ succ(c)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ x + a ' ≠ []], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z], [CSIMPSEQ : ⊦ # a ' = 0 +1], [CSIMPSEQ : ⊦ # (x + y) = # x + # y])]]