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Proof step 24 from proof compatval-put (conclusion)

⊦ compat(str, val, ty, st) ∧ fs .type ≠ void_type → compat(str, val, ty, st[r -- fs ', val0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
compat(str, val, ty, st) ⊦ fs .type = void_type, compat(str, val, ty, st[r -- fs ', val0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]